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.io.*;
import java.util.InputMismatchException;

/**
 * Generated by Contest helper plug-in
 * Actual solution is at the bottom
 */
public class Main {
    public static void main(String[] args) {
        InputReader in = new StreamInputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        run(in, out);
    }

    public static void run(InputReader in, PrintWriter out) {
        Solver solver = new TaskC();
        solver.solve(1, in, out);
        Exit.exit(in, out);
    }
}

abstract class InputReader {
    private boolean finished = false;

    public abstract int read();

    public int readInt() {
        return Integer.parseInt(readString());
    }

    public String readString() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuffer res = new StringBuffer();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    private boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    public boolean isFinished() {
        return finished;
    }

    public void setFinished(boolean finished) {
        this.finished = finished;
    }

    public abstract void close();
}

class StreamInputReader extends InputReader {
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar, numChars;

    public StreamInputReader(InputStream stream) {
        this.stream = stream;
    }

    public int read() {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    @Override
    public void close() {
        try {
            stream.close();
        } catch (IOException ignored) {
        }
    }
}

class Exit {
    private Exit() {
    }

    public static void exit(InputReader in, PrintWriter out) {
        in.setFinished(true);
        in.close();
        out.close();
    }
}

interface Solver {
    public void solve(int testNumber, InputReader in, PrintWriter out);
}

class TaskC implements Solver {
    public void solve(int testNumber, InputReader in, PrintWriter out) {
        int stringLength = in.readInt();
        int requests = in.readInt();
        int replaceCounter = 0;

        String str1 = in.readString();
        char[] str = str1.toCharArray();

        for(int i=0; i<stringLength; i++){
            if(i>0)
                if(str[i-1]=='.'&& str[i]=='.') replaceCounter++;
        }

        int index;
        char chr;

        for(int i=0; i<requests; i++){
            index = in.readInt();
            index--;
            chr = (char)in.read();

            if(chr=='.'){
                if(index<stringLength-1)
                    if(str[index]!='.' && str[index+1]=='.') replaceCounter++;

                if(index>0)
                    if(str[index]!='.' && str[index-1]=='.') replaceCounter++;
            }else{
                if(index<stringLength-1)
                    if(str[index]=='.' && str[index+1]=='.') replaceCounter--;

                if(index>0)
                    if(str[index]=='.' && str[index-1]=='.') replaceCounter--;
            }

            str[index] = chr;
            out.println(replaceCounter);
        }

    }
}