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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

22

2

When try to install dropbox from command line, I read the commands

cd ~ && wget -O - "https://www.dropbox.com/download?plat=lnx.x86_64" | tar xzf -

what does - mean here? is it previous directory?

command-line tar

share|improve this question

edited Jun 19 at 19:25

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

asked Jun 18 at 6:55

DummyHeadDummyHead

70344 silver badges2020 bronze badges

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  
  A bit of pedantry: the option you're asking about is f -. This might make finding the answer in the man pages a little easier.
  
  – studog
  Jun 18 at 19:01
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  It would be really helpful if you could explain what, precisely is unclear to you about the documentation of wget and tar. That way, the respective developers can improve their documentation so that future users don't stumble over the same problems again. In other words: be a hero and make the world a better place!
  
  – Jörg W Mittag
  Jun 18 at 20:07
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  It would be a nice default for tar to use stdin, but since it’s heritage is to work with tapes it did not do that. Instead it learned how to work with files (with the f option) and then follows the convention to have a - in place of the filename for stdin (In case of eXtract or stdout in case of Create)
  
  – eckes
  Jun 19 at 1:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @eckes In the case of GNU tar, stdin is the default unless $TAPE is set.
  
  – chrylis
  Jun 20 at 6:17
  

  
  
  
  

add a comment | 

22

2

When try to install dropbox from command line, I read the commands

cd ~ && wget -O - "https://www.dropbox.com/download?plat=lnx.x86_64" | tar xzf -

what does - mean here? is it previous directory?

command-line tar

share|improve this question

edited Jun 19 at 19:25

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

asked Jun 18 at 6:55

DummyHeadDummyHead

70344 silver badges2020 bronze badges

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  
  A bit of pedantry: the option you're asking about is f -. This might make finding the answer in the man pages a little easier.
  
  – studog
  Jun 18 at 19:01
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  It would be really helpful if you could explain what, precisely is unclear to you about the documentation of wget and tar. That way, the respective developers can improve their documentation so that future users don't stumble over the same problems again. In other words: be a hero and make the world a better place!
  
  – Jörg W Mittag
  Jun 18 at 20:07
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  It would be a nice default for tar to use stdin, but since it’s heritage is to work with tapes it did not do that. Instead it learned how to work with files (with the f option) and then follows the convention to have a - in place of the filename for stdin (In case of eXtract or stdout in case of Create)
  
  – eckes
  Jun 19 at 1:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @eckes In the case of GNU tar, stdin is the default unless $TAPE is set.
  
  – chrylis
  Jun 20 at 6:17
  

  
  
  
  

add a comment | 

When try to install dropbox from command line, I read the commands

cd ~ && wget -O - "https://www.dropbox.com/download?plat=lnx.x86_64" | tar xzf -

what does - mean here? is it previous directory?

command-line tar

share|improve this question

edited Jun 19 at 19:25

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

asked Jun 18 at 6:55

DummyHeadDummyHead

70344 silver badges2020 bronze badges

cd ~ && wget -O - "https://www.dropbox.com/download?plat=lnx.x86_64" | tar xzf -

share|improve this question

edited Jun 19 at 19:25

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

asked Jun 18 at 6:55

DummyHeadDummyHead

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share|improve this question

edited Jun 19 at 19:25

Eliah Kagan

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asked Jun 18 at 6:55

DummyHeadDummyHead

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edited Jun 19 at 19:25

Eliah Kagan

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edited Jun 19 at 19:25

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

asked Jun 18 at 6:55

DummyHeadDummyHead

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asked Jun 18 at 6:55

DummyHeadDummyHead

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3 Answers
3

active

oldest

votes

40

Some commands accept - in place of a filename, either:

• To write to standard output instead of to a named file. This is what the - argument passed to wget after -O is doing.


• To read from standard input instead of from a named file. This is what the - argument passed to tar after xzf.

The command you showed downloads an archive file with wget and unpacks it with tar. To achieve this, the output of wget is piped (|) to the input of tar. This is why wget writes to standard output instead of a file and tar reads from standard input instead of a file.

share|improve this answer

edited Jun 18 at 17:29

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd say "most commands" rather than just "some commands". I believe it's part of the POSIX option syntax specification.
  
  – Barmar
  Jun 18 at 15:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Barmar Yes and no: Guideline 13: For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named -.. So if - is an operand for a "file option" and it has a special meaning, then it must be that of stdin/stdout. The program is free to not treat as a special case and simply look for file -
  
  – Bakuriu
  Jun 19 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  (and typically in those cases omitting the option will be the same as using -)
  
  – Bakuriu
  Jun 19 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bakuriu I don't see how that refutes what I said. I suspect most commands take filename arguments, so they're supposed to follow that guideline. But maybe I should have qualified it as "most commands that operate on files".
  
  – Barmar
  Jun 19 at 19:31
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Barmar One way to follow the guideline is to decline to treat - specially at all, treating it as a literal filename. The only way to go against the guideline is to treat - specially but with some other meaning than the ones the guideline suggests.
  
  – Eliah Kagan
  Jun 19 at 19:39
  

  
  
  
  

 | 
show 3 more comments

12

That's just a filename that a lot of Unix programs interpret as "instead of actually opening a file, read from stdin (or write to stdout)."

That means reading from the input that gets streamed into the program; in your case, that's the output of wget.

share|improve this answer

edited Jun 18 at 19:58

Eliah Kagan

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answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

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New contributor

Marcus Müller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  Is it really a filename, though?
  
  – Eric Duminil
  Jun 18 at 17:01
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  @EricDuminil It's a filename in the sense that programs accept it in some places where they would otherwise only accept filenames, but you're right that it's not really about an actual name of a file in some directory.
  
  – JoL
  Jun 18 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @EricDuminil: In particular, for programs that follow this convention, you have to use ./- if you actually want to open a file named -.
  
  – Jörg W Mittag
  Jun 18 at 20:05
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @EricDuminil In fact: yes! It names a file; in the unix sense, stdin is a file descriptor, and - is the name you supply to use that. It's not a file name from the OS's perspective, though. Only from the user perspective.
  
  – Marcus Müller
  Jun 19 at 6:15
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @rexkogitans no. /dev/stdin is just a convenience file name offered by some kernel facility. There's usually no fopen("/dev/stdin", "r") happening anywhere when you write a program that chooses to use stdin; you just go ahead and use the extern FILE* stdin that your libc header provided you with; by the way, that is the integer file descriptor 0.
  
  – Marcus Müller
  Jun 19 at 6:46
  

  
  
  
  

 | 
show 5 more comments

5

The - argument to tar specifies that the archive should be read from stdin instead of a file. From the GNU tar manual:

If you use - as an archive-name, tar reads the archive from standard input (when listing or extracting files)

Other commands have the same behavior, and it is specified by the POSIX.1-2017 standard:

Guideline 13:

For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean
only standard input (or standard output when it is clear from context
that an output file is being specified) or a file named -.

share|improve this answer

answered Jun 18 at 22:01

TimTim

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add a comment | 

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40

Some commands accept - in place of a filename, either:

• To write to standard output instead of to a named file. This is what the - argument passed to wget after -O is doing.


• To read from standard input instead of from a named file. This is what the - argument passed to tar after xzf.

The command you showed downloads an archive file with wget and unpacks it with tar. To achieve this, the output of wget is piped (|) to the input of tar. This is why wget writes to standard output instead of a file and tar reads from standard input instead of a file.

share|improve this answer

edited Jun 18 at 17:29

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd say "most commands" rather than just "some commands". I believe it's part of the POSIX option syntax specification.
  
  – Barmar
  Jun 18 at 15:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Barmar Yes and no: Guideline 13: For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named -.. So if - is an operand for a "file option" and it has a special meaning, then it must be that of stdin/stdout. The program is free to not treat as a special case and simply look for file -
  
  – Bakuriu
  Jun 19 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  (and typically in those cases omitting the option will be the same as using -)
  
  – Bakuriu
  Jun 19 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bakuriu I don't see how that refutes what I said. I suspect most commands take filename arguments, so they're supposed to follow that guideline. But maybe I should have qualified it as "most commands that operate on files".
  
  – Barmar
  Jun 19 at 19:31
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Barmar One way to follow the guideline is to decline to treat - specially at all, treating it as a literal filename. The only way to go against the guideline is to treat - specially but with some other meaning than the ones the guideline suggests.
  
  – Eliah Kagan
  Jun 19 at 19:39
  

  
  
  
  

 | 
show 3 more comments

40

Some commands accept - in place of a filename, either:

• To write to standard output instead of to a named file. This is what the - argument passed to wget after -O is doing.


• To read from standard input instead of from a named file. This is what the - argument passed to tar after xzf.

The command you showed downloads an archive file with wget and unpacks it with tar. To achieve this, the output of wget is piped (|) to the input of tar. This is why wget writes to standard output instead of a file and tar reads from standard input instead of a file.

share|improve this answer

edited Jun 18 at 17:29

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'd say "most commands" rather than just "some commands". I believe it's part of the POSIX option syntax specification.
  
  – Barmar
  Jun 18 at 15:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @Barmar Yes and no: Guideline 13: For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean only standard input (or standard output when it is clear from context that an output file is being specified) or a file named -.. So if - is an operand for a "file option" and it has a special meaning, then it must be that of stdin/stdout. The program is free to not treat as a special case and simply look for file -
  
  – Bakuriu
  Jun 19 at 19:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  (and typically in those cases omitting the option will be the same as using -)
  
  – Bakuriu
  Jun 19 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bakuriu I don't see how that refutes what I said. I suspect most commands take filename arguments, so they're supposed to follow that guideline. But maybe I should have qualified it as "most commands that operate on files".
  
  – Barmar
  Jun 19 at 19:31
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Barmar One way to follow the guideline is to decline to treat - specially at all, treating it as a literal filename. The only way to go against the guideline is to treat - specially but with some other meaning than the ones the guideline suggests.
  
  – Eliah Kagan
  Jun 19 at 19:39
  

  
  
  
  

 | 
show 3 more comments

Some commands accept - in place of a filename, either:

• To write to standard output instead of to a named file. This is what the - argument passed to wget after -O is doing.


• To read from standard input instead of from a named file. This is what the - argument passed to tar after xzf.

The command you showed downloads an archive file with wget and unpacks it with tar. To achieve this, the output of wget is piped (|) to the input of tar. This is why wget writes to standard output instead of a file and tar reads from standard input instead of a file.

share|improve this answer

edited Jun 18 at 17:29

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

share|improve this answer

edited Jun 18 at 17:29

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

answered Jun 18 at 7:07

Eliah KaganEliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

12

That's just a filename that a lot of Unix programs interpret as "instead of actually opening a file, read from stdin (or write to stdout)."

That means reading from the input that gets streamed into the program; in your case, that's the output of wget.

share|improve this answer

edited Jun 18 at 19:58

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

22111 silver badge55 bronze badges

New contributor

Marcus Müller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  Is it really a filename, though?
  
  – Eric Duminil
  Jun 18 at 17:01
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  @EricDuminil It's a filename in the sense that programs accept it in some places where they would otherwise only accept filenames, but you're right that it's not really about an actual name of a file in some directory.
  
  – JoL
  Jun 18 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @EricDuminil: In particular, for programs that follow this convention, you have to use ./- if you actually want to open a file named -.
  
  – Jörg W Mittag
  Jun 18 at 20:05
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @EricDuminil In fact: yes! It names a file; in the unix sense, stdin is a file descriptor, and - is the name you supply to use that. It's not a file name from the OS's perspective, though. Only from the user perspective.
  
  – Marcus Müller
  Jun 19 at 6:15
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @rexkogitans no. /dev/stdin is just a convenience file name offered by some kernel facility. There's usually no fopen("/dev/stdin", "r") happening anywhere when you write a program that chooses to use stdin; you just go ahead and use the extern FILE* stdin that your libc header provided you with; by the way, that is the integer file descriptor 0.
  
  – Marcus Müller
  Jun 19 at 6:46
  

  
  
  
  

 | 
show 5 more comments

12

That's just a filename that a lot of Unix programs interpret as "instead of actually opening a file, read from stdin (or write to stdout)."

That means reading from the input that gets streamed into the program; in your case, that's the output of wget.

share|improve this answer

edited Jun 18 at 19:58

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

22111 silver badge55 bronze badges

New contributor

Marcus Müller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  
  Is it really a filename, though?
  
  – Eric Duminil
  Jun 18 at 17:01
  

  
  
  
  


• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  
  @EricDuminil It's a filename in the sense that programs accept it in some places where they would otherwise only accept filenames, but you're right that it's not really about an actual name of a file in some directory.
  
  – JoL
  Jun 18 at 17:55
  
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  @EricDuminil: In particular, for programs that follow this convention, you have to use ./- if you actually want to open a file named -.
  
  – Jörg W Mittag
  Jun 18 at 20:05
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  @EricDuminil In fact: yes! It names a file; in the unix sense, stdin is a file descriptor, and - is the name you supply to use that. It's not a file name from the OS's perspective, though. Only from the user perspective.
  
  – Marcus Müller
  Jun 19 at 6:15
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @rexkogitans no. /dev/stdin is just a convenience file name offered by some kernel facility. There's usually no fopen("/dev/stdin", "r") happening anywhere when you write a program that chooses to use stdin; you just go ahead and use the extern FILE* stdin that your libc header provided you with; by the way, that is the integer file descriptor 0.
  
  – Marcus Müller
  Jun 19 at 6:46
  

  
  
  
  

 | 
show 5 more comments

That's just a filename that a lot of Unix programs interpret as "instead of actually opening a file, read from stdin (or write to stdout)."

That means reading from the input that gets streamed into the program; in your case, that's the output of wget.

share|improve this answer

edited Jun 18 at 19:58

Eliah Kagan

85.4k2222 gold badges239239 silver badges380380 bronze badges

answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

22111 silver badge55 bronze badges

New contributor

Marcus Müller is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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edited Jun 18 at 19:58

Eliah Kagan

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answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

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edited Jun 18 at 19:58

Eliah Kagan

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edited Jun 18 at 19:58

Eliah Kagan

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answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

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answered Jun 18 at 7:07

Marcus MüllerMarcus Müller

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5

The - argument to tar specifies that the archive should be read from stdin instead of a file. From the GNU tar manual:

If you use - as an archive-name, tar reads the archive from standard input (when listing or extracting files)

Other commands have the same behavior, and it is specified by the POSIX.1-2017 standard:

Guideline 13:

For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean
only standard input (or standard output when it is clear from context
that an output file is being specified) or a file named -.

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answered Jun 18 at 22:01

TimTim

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5

The - argument to tar specifies that the archive should be read from stdin instead of a file. From the GNU tar manual:

If you use - as an archive-name, tar reads the archive from standard input (when listing or extracting files)

Other commands have the same behavior, and it is specified by the POSIX.1-2017 standard:

Guideline 13:

For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean
only standard input (or standard output when it is clear from context
that an output file is being specified) or a file named -.

share|improve this answer

answered Jun 18 at 22:01

TimTim

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Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

The - argument to tar specifies that the archive should be read from stdin instead of a file. From the GNU tar manual:

If you use - as an archive-name, tar reads the archive from standard input (when listing or extracting files)

Other commands have the same behavior, and it is specified by the POSIX.1-2017 standard:

Guideline 13:

For utilities that use operands to represent files to be opened for either reading or writing, the '-' operand should be used to mean
only standard input (or standard output when it is clear from context
that an output file is being specified) or a file named -.

share|improve this answer

answered Jun 18 at 22:01

TimTim

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share|improve this answer

answered Jun 18 at 22:01

TimTim

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answered Jun 18 at 22:01

TimTim

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answered Jun 18 at 22:01

TimTim

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