ÿØÿàJFIFÿþ ÿÛC       ÿÛC ÿÀÿÄÿÄ"#QrÿÄÿÄ&1!A"2qQaáÿÚ ?Øy,æ/3JæݹÈ߲؋5êXw²±ÉyˆR”¾I0ó2—PI¾IÌÚiMö¯–þrìN&"KgX:Šíµ•nTJnLK„…@!‰-ý ùúmë;ºgµŒ&ó±hw’¯Õ@”Ü— 9ñ-ë.²1<yà‚¹ïQÐU„ہ?.’¦èûbß±©Ö«Âw*VŒ) `$‰bØԟ’ëXÖ-ËTÜíGÚ3ð«g Ÿ§¯—Jx„–’U/ÂÅv_s(Hÿ@TñJÑãõçn­‚!ÈgfbÓc­:él[ðQe 9ÀPLbÃãCµm[5¿ç'ªjglå‡Ûí_§Úõl-;"PkÞÞÁQâ¼_Ñ^¢SŸx?"¸¦ùY騐ÒOÈ q’`~~ÚtËU¹CڒêV  I1Áß_ÿÙfrom distutils.dep_util import newer_group # yes, this is was almost entirely copy-pasted from # 'newer_pairwise()', this is just another convenience # function. def newer_pairwise_group(sources_groups, targets): """Walk both arguments in parallel, testing if each source group is newer than its corresponding target. Returns a pair of lists (sources_groups, targets) where sources is newer than target, according to the semantics of 'newer_group()'. """ if len(sources_groups) != len(targets): raise ValueError( "'sources_group' and 'targets' must be the same length") # build a pair of lists (sources_groups, targets) where source is newer n_sources = [] n_targets = [] for i in range(len(sources_groups)): if newer_group(sources_groups[i], targets[i]): n_sources.append(sources_groups[i]) n_targets.append(targets[i]) return n_sources, n_targets