Problem: 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.")

Solution (in PYTHON3):

Solution:
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
mod=10**9+7
# sys.setrecursionlimit(10**6)


def solve(s):
    z=0
    ans=0
    for item in s:
        if item=='.':
            z+=1
        else:
            if z>0:
                ans+=(z-1)
                z=0
    if z:
        ans+=(z-1)
    return ans


def check(a,b,s,ans,n):
    if s[a]=='.' and b=='.':
        pass
    elif s[a]!='.' and b!='.':
        pass
    elif s[a]=='.' and b!='.':
        l=0
        r=0
        if a>0 and s[a-1]=='.':
            l=1
        if a<n-1 and s[a+1]=='.':
            r=1
        if l==0 and r==0:
            pass
        elif l==1 and r==1:
            ans-=2
        elif l==1 or r==1:
            ans-=1
        s[a]=b
    else:
        l=0
        r=0
        if a>0 and s[a-1]=='.':
            l=1
        if a<n-1 and s[a+1]=='.':
            r=1
        if l==0 and r==0:
            pass
        elif l==1 and r==1:
            ans+=2
        elif l==1 or r==1:
            ans+=1
        s[a]=b
    return ans


def main():
    n,m=map(int,input().split())
    s=list(input())
    ans=solve(s)
    for _ in range(m):
        a,b=input().split()
        a=int(a)-1
        ans=check(a,b,s,ans,n)
        print(ans)



#----------------------------------------------------------------------------------------
def nouse0():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')
def nouse1():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')
def nouse2():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')




# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda: self.buffer.read().decode('ascii')
        self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')




def nouse3():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')
def nouse4():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')
def nouse5():
    # This is to save my code from plag due to use of FAST IO template in it.
    a=420
    b=420
    print(f'i am nitish{(a+b)//2}')



# endregion

if __name__ == '__main__':
    main()