CF570C Replacement

Daniel has a string $ s $ , consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacement as the following sequence of steps: find a substring ".." (two consecutive periods) in string $ s $ , of all occurrences of the substring let's choose the first one, and replace this substring with string ".". In other words, during the replacement operation, the first two consecutive periods are replaced by one. If string $ s $ contains no two consecutive periods, then nothing happens. Let's define $ f(s) $ as the minimum number of operations of replacement to perform, so that the string does not have any two consecutive periods left. You need to process $ m $ queries, the $ i $ -th results in that the character at position $ x_{i} $ ( $ 1

The first line contains two integers $ n $ and $ m $ ( $ 1

Print $ m $ numbers, one per line, the $ i $ -th of these numbers must be equal to the value of $ f(s) $ after performing the $ i $ -th assignment.

Note to the first sample test (replaced periods are enclosed in square brackets). The original string is ".b..bz....". - after the first query $ f( $ hb..bz.... $ ) $ = 4 ("hb\[..\]bz...." $ → $ "hb.bz\[..\].." $ → $ "hb.bz\[..\]." $ → $ "hb.bz\[..\]" $ → $ "hb.bz.") - after the second query $ f( $ hbс.bz.... $ ) $ = 3 ("hbс.bz\[..\].." $ → $ "hbс.bz\[..\]." $ → $ "hbс.bz\[..\]" $ → $ "hbс.bz.") - after the third query $ f( $ hbс.bz..f. $ ) $ = 1 ("hbс.bz\[..\]f." $ → $ "hbс.bz.f.") Note to the second sample test. The original string is ".cc.". - after the first query: $ f( $ ..c. $ ) $ = 1 ("\[..\]c." $ → $ ".c.") - after the second query: $ f( $ .... $ ) $ = 3 ("\[..\].." $ → $ "\[..\]." $ → $ "\[..\]" $ → $ ".") - after the third query: $ f( $ .a.. $ ) $ = 1 (".a\[..\]" $ → $ ".a.") - after the fourth query: $ f( $ aa.. $ ) $ = 1 ("aa\[..\]" $ → $ "aa.")