Solve in Python:
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.


-----Examples-----
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...
import sys; sys.setrecursionlimit(1000000)
def solve():
    n, m, = rv()
    s = list(input())
    res = [0] * m
    #replace dot:
    #dot had nothing on left or right: nothing changes
    #dot had one on left or right: -1
    #dot had two on left or right: -2

    #replace char:
    #if had two chars on left and right: 0
    # if had one char and one dot: +1
    # if had two dots: +2
    helper = list()
    for i in range(n):
        if s[i] == '.':
            if i == 0:
                helper.append(1)
            else:
                if s[i-1] == '.':
                    helper[-1] += 1
                else:
                    helper.append(1)
    initval = 0
    for val in helper:
        initval += val - 1
    for query in range(m):
        index, replace = input().split()
        index = int(index) - 1
        if (s[index] == '.' and replace == '.') or (s[index] != '.' and replace != '.'):
            res[query] = initval
        else:
            sidedots = 0
            if index > 0:
                if s[index - 1] == '.': sidedots+=1
            if index < n - 1:
                if s[index + 1] == '.': sidedots+=1
            if s[index] == '.':
                res[query] = initval - sidedots
                initval -= sidedots
            else:
                res[query] = initval + sidedots
                initval += sidedots
        s[index] = replace
    print('\n'.join(map(str, res)))




def rv(): return list(map(int, input().split()))
def rl(n): return [list(map(int, input().split())) for _ in range(n)]
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()