problem running a python file

[alireza-saberi]

I am running a python file on shell by the following python ./tools/make-firefox-meta.py ../adaware.firefox_debug/

but when I put it into gulp-run, with following as a task

gulp.task("python-bsff", ["make-pre-bsff"], function(){
    return run("python ./tools/make-firefox-meta.py ../adaware.firefox_debug/").exec();
}); 

The output is weird

[09:59:03] Error: Command `python ./tools/make-firefox-meta.py ../adaware.firefox_debug/` exited with code 1
    at ChildProcess.handleSubShellExit (C:\Users\alan.saberi\dev\AdAwareAdBlockCore\node_modules\gulp-run\command.js:166:13)
    at ChildProcess.g (events.js:291:16)
    at emitTwo (events.js:106:13)
    at ChildProcess.emit (events.js:191:7)
    at Process.ChildProcess._handle.onexit (internal/child_process.js:215:12)
stream.js:74
      throw er; // Unhandled stream error in pipe.
      ^

Error: ENOENT: no such file or directory, stat 'C:\Users\alan.saberi\dev\adaware.firefox_debug\_locales\bg\messages.json'
    at Error (native)

Can you explain to me please, what is wrong with gulp-run and how can I pass this error?

[alireza-saberi]

About the error: Error: ENOENT: no such file or directory, stat 'C:\Users\alan.saberi\dev\adaware.firefox_debug\_locales\bg\messages.json' at Error (native)

before running this file messages.json exists. and normal shell command works on it, but gulp-run is messing up somewhere