Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacementas 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 xi (1 ≤ xi ≤ n) of string s is assigned value ci. After each operation you have to calculate and output the value of f(s).
Help Daniel to process all queries.
The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) the length of the string and the number of queries.
The second line contains string s, consisting of n lowercase English letters and period signs.
The following m lines contain the descriptions of queries. The i-th line contains integer xi and ci (1 ≤ xi ≤ n, ci — a lowercas English letter or a period sign), describing the query of assigning symbol ci to position xi.
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.
10 3 .b..bz.... 1 h 3 c 9 f
4 3 1
4 4 .cc. 2 . 3 . 2 a 1 a
1 3 1 1
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.")
题目大意：定义对于字符串s的一次操作为，在s中任选两个连续的句点，并把它们替换成一个句点。定义$f\left( s\right)$为使字符串s不再有两个连续句点的所需的最小操作数。现在有一个长度为n的字符串s。有m次查询，第i次查询把s的第$x_i$个字符替换成$c_i$（对之后的查询也有效），并要求输出$f\left( s\right)$。
设s中的句点总数为tot，连续句点段数为cnt，显然有$f\left( s\right) =tot-cnt$。只需在每次查询时维护这两个值就行了。
代码在此。
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