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 JAVA):

Solution:
import java.awt.Point;
import java.awt.Rectangle;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.ObjectInputStream.GetField;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.security.KeyStore.Entry;
import java.text.DecimalFormat;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collections;
import java.util.Comparator;
import java.util.Deque;
import java.util.Formatter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Solution {
    static FastReader in = new FastReader(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(System.out);
    static boolean file = false;
    static final int maxn = (int)2e3+111;
    static int inf = (int)1e9;
    static int n,m;
    static int a[] = new int [maxn];
    static int cnt[] = new int [maxn];
    static StringBuilder s = new StringBuilder();
    
    private static void solve() throws Exception {
        n = in.nextInt();
        m = in.nextInt();
        s.append(in.nextToken());
        int cnt = 0;
        for (int i=0; i<n-1; i++) {
            if (s.charAt(i)=='.' && s.charAt(i+1)=='.') 
                cnt++;
        }
        
        for (int i=1; i<=m; i++) {
            int num = in.nextInt();
            char val = in.nextToken().charAt(0);
            num--;
            if (val=='.') {
                if (s.charAt(num)!='.') {
                    if (num-1>=0 && s.charAt(num-1)=='.') cnt++;
                    if (num+1<n && s.charAt(num+1)=='.') cnt++;
                }
            } else {
                if (s.charAt(num)=='.') {
                    if (num-1>=0 && s.charAt(num-1)=='.') cnt--;
                    if (num+1<n && s.charAt(num+1)=='.') cnt--;
                }
            }
            s.setCharAt(num, val);
            out.println(cnt);
        }
    }

    public static void main (String [] args) throws Exception {
        
        if (file) {
            in = new FastReader(new BufferedReader(new FileReader("input.txt")));
            out = new PrintWriter ("output.txt");
        }
       
        solve();
        out.close();
    }

}

class PairInPair implements Comparable<PairInPair> {
    Pair p;
    int id;
    
    public PairInPair (Pair p, int id) {
        this.p = p;
        this.id = id;
    }

    @Override
    public int compareTo(PairInPair pair) {
        if (p.x>pair.p.x) return 1;
        else if (p.x==pair.p.x) {
            if (p.y>pair.p.y) return 1;
            else if (p.y==pair.p.y) return 0;
            else return -1;
        }
        else return -1;
    }
}

class Pair implements Comparable<Pair> {
    long x,y;
    public Pair(long x, long y) {
        this.x = x;
        this.y = y;
    }

    

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + (int) (x ^ (x >>> 32));
        result = prime * result + (int) (y ^ (y >>> 32));
        return result;
    }



    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Pair other = (Pair) obj;
        if (x != other.x)
            return false;
        if (y != other.y)
            return false;
        return true;
    }

    @Override
    public int compareTo(Pair p) {
        if (x>p.x)return 1;
        else if (x==p.x) {
            if (y>p.y) return 1;
            else if (y==p.y) return 0;
            else return -1;
        }
        else return -1;
    }
}

class FastReader {
    BufferedReader bf;
    StringTokenizer tk = null;

    public FastReader(BufferedReader bf) {
        this.bf = bf;
    }
    
    public String nextToken () throws Exception {
        if (tk==null || !tk.hasMoreTokens()) {
            tk = new StringTokenizer(bf.readLine());
        }
        return tk.nextToken();
    }
    
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }
    
    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }
    
    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}