题目描述

Description

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 ≤ $x_i$ ≤ n) of string s is assigned value $c_i$. After each operation you have to calculate and output the value of f(s).

Help Daniel to process all queries.

Input

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 $x_i$ and $c_i$ (1 ≤ $x_i$ ≤ n, $c_i$ — a lowercas English letter or a period sign), describing the query of assigning symbol $c_i$ to position $x_i$.

Output

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.

Sample test(s)

input
10 3
.b..bz….
1 h
3 c
9 f

output
4
3
1

input
4 4
.cc.
2 .
3 .
2 a
1 a

output
1
3
1
1

Note

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.”)

题目链接

http://codeforces.com/problemset/problem/570/C

解题思路

一开始以为是一道多难的题，竟然如此的简单。比赛还剩不到10分钟的时候才想到，匆忙中敲了一发，竟然WA了，也没时间检查哪里错了。赛后看了一下别人的代码，才发现自己的代码写Low了。

AC代码

171 ms/100 KB/GNU C++

#include <cstdio>

char str[300005], ch[5];
int main()
{
    int n, m, x;
    while (~scanf("%d%d", &n, &m))
    {
        int cnt = 0;
        scanf("%s", str + 1);
        for (int i = 1; i < n; i++)
        {
            cnt += str[i] == '.' && str[i + 1] == '.';
        }
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%s", &x, ch);
            cnt -= str[x - 1] == '.' && str[x] == '.';
            cnt -= str[x + 1] == '.' && str[x] == '.';
            str[x] = ch[0];
            cnt += str[x - 1] == '.' && str[x] == '.';
            cnt += str[x + 1] == '.' && str[x] == '.';
            printf("%d\n", cnt);
        }
    }
    return 0;
}