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 PYTHON):

Solution:
# -*- coding: utf-8 -*-

# Universidade Federal de Campina Grande
# Aluno: Diego Adolfo Silva de Araújo
# Matricula: 113210090
# Disciplina: Algoritmos Avançados

# Codeforces
# Problem: 570C - C. Replacement
# Time limit per test: 2 second
# Memory limit per test: 256 megabytes
# Input: standard input
# Output: standard output

def replacement(words):
    value, number = 0, 0
    for character in words:
       if character == '.':
          number += 1
       elif number > 0:
	  value += number - 1
          number = 0
    
    if number > 0:
       value += number - 1
    return value

empty_str = [' ']
# the lenght of the string and number of queries
n, m = map(int, raw_input().split())
# the string s, consisting of n lowercase English letters and period signs (characters '.')
string = empty_str + map(None, raw_input()) + empty_str
value = replacement(string[1:n+1])
for query in xrange(m):
    # a position and a lowercase English letter or period sing, describing the query
    position, symbol = raw_input().split()
    position = int(position)
	
    if symbol != '.' and string[position] == '.':
       if string[position - 1] == '.' and string[position + 1] == '.':
	  value -= 2
       elif string[position - 1] == '.' or string[position + 1] == '.':
	  value -= 1
    elif symbol == '.' and string[position] != '.':
       if string[position - 1] == '.' and string[position + 1] == '.':
          value += 2
       elif string[position - 1] == '.' or string[position + 1] == '.':
          value += 1

    string[position] = symbol 
    # print numbers to the value of replacement(string) after performing the i-th assignment
    print value