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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int N = 3e5 + 100;
int tree[N << 2];
void upd(int node, int st, int ed, int idx, int val) {
  if (st == ed) {
    tree[node] += val;
    return;
  }
  int mid = st + ed >> 1;
  if (idx <= mid) {
    upd(node << 1, st, mid, idx, val);
  } else {
    upd((node << 1) | 1, mid + 1, ed, idx, val);
  }
  tree[node] = tree[node << 1] + tree[(node << 1) | 1];
}
int query(int node, int st, int ed, int l, int r) {
  if (l > ed || r < st) {
    return 0;
  }
  if (l <= st && r >= ed) {
    return tree[node];
  }
  int mid = st + ed >> 1;
  int lft = query(node << 1, st, mid, l, r);
  int rgt = query((node << 1) | 1, mid + 1, ed, l, r);
  return lft + rgt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  string s;
  cin >> s;
  s = '#' + s;
  for (int i = 1; i <= n; i++) {
    if (s[i] == '.') {
      upd(1, 1, n, i, 1);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] != '.') continue;
    int j = i;
    while (i <= n && s[i] == '.') i++;
    ans += i - j - 1;
  }
  while (q--) {
    int tmp = ans;
    int pos;
    char ch;
    cin >> pos >> ch;
    if (s[pos] == '.') {
      if (ch == '.') {
        cout << tmp << '\n';
        continue;
      }
      upd(1, 1, n, pos, -1);
      s[pos] = ch;
      int lo = pos, hi = n + 1;
      while (hi - lo > 1) {
        int mid = lo + hi >> 1;
        if (query(1, 1, n, pos + 1, mid) == mid - pos)
          lo = mid;
        else
          hi = mid;
      }
      int R = lo;
      lo = 0, hi = pos;
      while (hi - lo > 1) {
        int mid = lo + hi >> 1;
        if (query(1, 1, n, mid, pos - 1) == pos - mid)
          hi = mid;
        else
          lo = mid;
      }
      int L = hi;
      ans -= R - L;
      ans += max(0, R - pos - 1) + max(0, pos - L - 1);
      cout << ans << '\n';
    } else {
      if (ch != '.') {
        cout << tmp << '\n';
        continue;
      }
      upd(1, 1, n, pos, 1);
      s[pos] = '.';
      int lo = pos, hi = n + 1;
      while (hi - lo > 1) {
        int mid = lo + hi >> 1;
        if (query(1, 1, n, pos + 1, mid) == mid - pos)
          lo = mid;
        else
          hi = mid;
      }
      int R = lo;
      lo = 0, hi = pos;
      while (hi - lo > 1) {
        int mid = lo + hi >> 1;
        if (query(1, 1, n, mid, pos - 1) == pos - mid)
          hi = mid;
        else
          lo = mid;
      }
      int L = hi;
      ans -= max(0, R - pos - 1) + max(0, pos - L - 1);
      ans += R - L;
      cout << ans << '\n';
    }
  }
  return 0;
}