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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool secdes(const pair<long long int, long long int>& a,
            const pair<long long int, long long int>& b) {
  if (a.second == b.second) {
    return a.first > b.first;
  } else {
    return a.second > b.second;
  }
}
bool sir(const pair<long long int, long long int>& a,
         const pair<long long int, long long int>& b) {
  if (a.first == b.first) {
    return a.second > b.second;
  } else {
    return a.first > b.first;
  }
}
bool bysec(const pair<long long int, long long int>& a,
           const pair<long long int, long long int>& b) {
  if (a.second == b.second) {
    return a.first < b.first;
  } else {
    return a.second < b.second;
  }
}
const long long int no = 1e6 + 5;
const long long int modulo = 1e9 + 7;
const long long int inf = 1e18;
const long long int N = 2001;
long long int ar[no + 10], br[no + 10], cr[no + 10], mat[N][N];
long long int dx[] = {-1, 0, 1, 0};
long long int dy[] = {0, 1, 0, -1};
char dc[] = {'L', 'D', 'R', 'U'};
long long int mul(long long int x, long long int y, long long int mod) {
  return ((x % mod) * (y % mod)) % mod;
}
long long int powwmod(long long int x, long long int y, long long int mod) {
  long long int res = 1;
  while (y) {
    if (y & 1) {
      y--;
      res = mul(res, x, mod);
      res %= mod;
    } else {
      y /= 2;
      x = mul(x, x, mod);
      x %= mod;
    }
  }
  return res % mod;
}
long long int poww(long long int x, long long int y) {
  long long int res = 1;
  while (y) {
    if (y % 2) {
      y--;
      res = res * x;
    } else {
      y /= 2;
      x = x * x;
    }
  }
  return res;
}
long long int div(long long int x, long long int y, long long int mod) {
  return (x * powwmod(y, mod - 2, modulo)) % mod;
}
long long int add(long long int x, long long int y, long long int mod) {
  return (x % mod + y % mod) % mod;
}
bool comp(long long int x, long long int y) { return (x < y); }
long long int gcd(long long int x, long long int y) {
  if (y == 0)
    return x;
  else
    return gcd(y, x % y);
}
long long int lcm(long long int x, long long int y) {
  return (x * y) / gcd(x, y);
}
vector<long long int> adj[no];
long long int used[no], color[no], weight[no];
long long int ans1 = 0, ans2 = 0;
void solve() {
  long long int n = 0, m = 0, a = 0, b = 0, c = 0, x = 0, y = 0, z = 0, w = 0,
                k = 0;
  cin >> n >> m;
  string s;
  cin >> s;
  s += '!';
  s = '!' + s;
  n++;
  for (long long int i = 1; i < n; i++) {
    if (s[i] == '.' && s[i] == s[i + 1]) {
      x++;
    } else if (s[i] == '.' && s[i] != s[i + 1]) {
      y += x;
      x = 0;
    }
  }
  while (m--) {
    cin >> x;
    char ch;
    cin >> ch;
    if (s[x] == '.' && ch != '.') {
      if (s[x - 1] == '.' && s[x + 1] == '.')
        y -= 2;
      else if (s[x - 1] == '.' || s[x + 1] == '.')
        y--;
    } else if (s[x] != '.' && ch == '.') {
      if (s[x - 1] == '.' && s[x + 1] == '.')
        y += 2;
      else if (s[x - 1] == '.' || s[x + 1] == '.')
        y++;
    }
    s[x] = ch;
    cout << y << "\n";
  }
}
inline void runn() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    solve();
  }
}