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 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.

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 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.

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 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.")
#!/usr/bin/env python3
"""
Codeforces Round #316 (Div. 2)

Problem 570 C. Replacement

@author yamaton
@date 2015-08-13
"""

import itertools as it
import bisect
import sys


DOT = '.'
BAR = '|'

def solve(s, pairs):
    # xs is called with 1-based index. 
    n = len(s)
    # 0-th and (n+1)-th elements are guards.
    xs = [BAR] + [DOT if c == DOT else BAR for c in s] + [BAR]
    idx_char_pairs = [(int(x), DOT if c == DOT else BAR) for (x, c) in pairs]  # 1-based index
    counts = [sum(1 for _ in iterable) for c, iterable in it.groupby(xs) if c == DOT]
    total = sum(counts) - len(counts)
    n = len(xs)
    # print_stderr(''.join(xs), total)

    result = []
    for idx, c in idx_char_pairs:
        # print_stderr('At idx {}: {} -> {}'.format(idx, xs[idx], c))
        if xs[idx] == c:                          # no change
            pass
        elif xs[idx] == DOT:     # . -> |
            xs[idx] = BAR
            if xs[idx - 1] ==  xs[idx + 1] == DOT:  
                #  ... -> .|.
                total -= 2
            elif xs[idx - 1] == DOT or xs[idx + 1] == DOT:  
                #  |.. -> ||.
                total -= 1
            else:
                # |.|  ->  |||
                pass

        elif xs[idx] == BAR:       # | -> .
            xs[idx] = DOT
            if xs[idx - 1] == xs[idx + 1] == DOT:
                # .|.  -> ...
                total += 2
            elif xs[idx - 1] == DOT or xs[idx + 1] == DOT:
                # .|| -> ..|
                total += 1
            else:
                # ||| -> |.|
                pass

        result.append(total)
        # print_stderr(''.join(xs), total)

    return result



# -- Too slow
def solve_slow(s, pairs):
    idx_char_pairs = [(int(x) - 1, '.' if c == '.' else 'x') for (x, c) in pairs]
    ss = list(s)

    result = []
    for (idx, c) in idx_char_pairs:
        ss[idx] = c
        dot_stepss = [sum(1 for _ in iterable) for (cc, iterable) in it.groupby(ss) if cc == '.']
        out = sum(cnt - 1 for cnt in dot_stepss)
        result.append(out)
    return result


def print_stderr(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)


def main():
    [n, m] = [int(i) for i in input().strip().split()]
    s = input().strip()
    assert len(s) == n
    xc_pairs = [input().strip().split() for _ in range(m)]

    # print_stderr(s)
    # print_stderr(xc_pairs)
    result = solve(s, xc_pairs)
    for x in result:
        print(x)


if __name__ == '__main__':
    main()
