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:
//package ;

import java.util.ArrayList;
import java.util.Scanner;
import java.io.*;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class TaskC {
    public static void main(String[] args){
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        /*long start = System.currentTimeMillis();
        for(int i = 0; i < 300000; i++){
            out.println(i);
        }
        out.println("takes " + (System.currentTimeMillis() - start));
        out.close();*/
        try {
            new TaskC().solve(1,sc, out);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /** call this method to initialize reader for InputStream */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input) );
            tokenizer = new StringTokenizer("");
        }

        /** get next word */
        static String next() throws IOException {
            while ( ! tokenizer.hasMoreTokens() ) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine() );
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt( next() );
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble( next() );
        }
    }
    public void solve(int testNumber, MyScanner in, PrintWriter out) throws IOException {
        Reader reader = new TaskC.Reader();
        reader.init(System.in);
        int n = reader.nextInt();
        int m = reader.nextInt();
        String s = reader.next();
        TreeSet<Seg> segs = new TreeSet<Seg>();
        boolean running = false;
        int curStart = 0;
        int ans = 0;
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) == '.'){
                if(!running){
                    curStart = i+1;
                    running = true;
                }
            }else if(running){
                running = false;
                segs.add(new Seg(curStart, i));
                ans += (i-curStart);
            }
        }
        if(running) {
            segs.add(new Seg(curStart, n));
            ans+=(n-curStart);
        }
        segs.add(new Seg(-100,-100));
        segs.add(new Seg(Integer.MAX_VALUE,Integer.MAX_VALUE));
        for(int q = 0; q < m; q++){
            int idx = reader.nextInt();
            String x = reader.next();
            Seg potential = new Seg(idx,idx);
            if(".".equals(x)){
                if(segs.contains(potential)) {
                    out.println(ans);
                    continue;
                }
                Seg c = segs.ceiling(potential);
                Seg f = segs.floor(potential);
                if(f.end == idx-1 && c.start == idx+1){
                    segs.remove(c);
                    segs.remove(f);
                    ans -= c.end-c.start;
                    ans -= f.end-f.start;
                    segs.add(new Seg(f.start, c.end));
                    ans += c.end -f.start;
                }else if(f.end == idx-1){
                    segs.remove(f);
                    ans -= f.end-f.start;
                    segs.add(new Seg(f.start, idx));
                    ans += idx -f.start;
                }else if(c.start == idx+1){
                    segs.remove(c);
                    ans -= c.end - c.start;
                    segs.add(new Seg(idx, c.end));
                    ans += c.end - idx;
                }else{
                    segs.add(potential);
                }
            }else{
                if(!segs.contains(potential)){
                    out.println(ans);
                    continue;
                }
                Seg ss = segs.ceiling(potential);
                ans -= ss.end - ss.start;
                segs.remove(ss);
                if(ss.start < idx){
                    Seg left = new Seg(ss.start, idx-1);
                    segs.add(left);
                    ans+=left.end-left.start;
                }
                if(ss.end > idx){
                    Seg right = new Seg(idx+1, ss.end);
                    segs.add(right);
                    ans+=right.end-right.start;
                }
            }
            out.println(ans);
        }
        out.close();
    }

    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }

    private class Seg implements Comparable<Seg>{
        int start, end;

        public Seg(int start, int end) {
            this.end = end;
            this.start = start;
        }

        @Override
        public int compareTo(Seg o) {
            if(end < o.start) return -1;
            if(start > o.end) return 1;
            return 0;
        }
    }
}