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View submission: /r/xkcd is free?
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Where does the link in that one lead?
Comment by Wyboth at 03/10/2014 at 14:50 UTC
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I think it was this,[1] but I'm not sure.
1: http://www.reddit.com/r/anonymous123421/comments/1w8aie/petition_to_reinstate_uwyboth_as_a_mod_of_rxkcd/cezp63m?context=3