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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.StringTokenizer;

public class Problem3 {

    private static final boolean SUBMIT = true;
    private final BufferedReader in;
    private final PrintWriter out;
    private StringTokenizer tokenizer;

    private Problem3(BufferedReader in, PrintWriter out) {
        this.in = in;
        this.out = out;
        this.tokenizer = new StringTokenizer("");
    }

    private static Problem3 createProblem() {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        return new Problem3(in, out);
    }

    public static void main(String[] args) {
        try {
            Problem3 solution = createProblem();
            if (SUBMIT) {
                solution.solve();
            } else {
                solution.solveMany();
            }
            solution.close();
        } catch (Exception e) {
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }

    public void solveMany() throws Exception {
        int n = readInt();
        for (int i = 0; i < n; i++) {
            timeSolve();
        }
    }

    public void timeSolve() throws Exception {
        long currentTimeMillis = System.currentTimeMillis();
        solve();
        out.flush();
        double timeTaken = System.currentTimeMillis() - currentTimeMillis;
        out.println("time taken: " + timeTaken / 1000);
        out.println("");
    }

    private String readString() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(in.readLine());
        }
        return tokenizer.nextToken();
    }

    private char readChar() throws IOException {
        return readString().toCharArray()[0];
    }

    private int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    private void close() throws IOException {
        in.close();
        out.close();
    }

    private void solve() throws Exception {
        int n = readInt();
        int m = readInt();
        char[] word = readString().toCharArray();
        boolean[] isPeriod = new boolean[n+2];
        boolean[] isAfterPeriod = new boolean[n+2];
        boolean[] isBeforePeriod = new boolean[n+2];
        int streak = 0;
        int f = 0;
        for (int i = 0; i < n; i++) {
            if (word[i] == '.') {
                isPeriod[i+1] = true;
                isAfterPeriod[i+2] = true;
                isBeforePeriod[i] = true;
                streak += 1;
            } else {
                f += Math.max(streak - 1, 0);
                streak = 0;
            }
        }
        f += Math.max(streak - 1, 0);
        for (int i = 0; i < m; i++) {
            int index = readInt();
            boolean period = readChar() == '.';
            if (period) {
                if (!isPeriod[index]) {
                    if (isAfterPeriod[index]) {
                        f += 1;
                    }
                    if (isBeforePeriod[index]) {
                        f += 1;
                    }
                    isPeriod[index] = true;
                    isAfterPeriod[index+1] = true;
                    isBeforePeriod[index-1] = true;
                }
            } else {
                if (isPeriod[index]) {
                    if (isAfterPeriod[index]) {
                        f -= 1;
                    }
                    if (isBeforePeriod[index]) {
                        f -= 1;
                    }
                    isPeriod[index] = false;
                    isAfterPeriod[index+1] = false;
                    isBeforePeriod[index-1] = false;
                }
            }
            out.println(f);
        }
    }

    private void solve2() throws Exception {
        int n = readInt();
        int m = readInt();
        char[] word = readString().toCharArray();
        List<Integer> beginIndices = new LinkedList<>();
        List<Integer> endIndices = new LinkedList<>();
        boolean inPeriods = false;
        for (int i = 0; i < n; i++) {
            char c = word[i];
            if (c == '.') {
                if (!inPeriods) {
                    beginIndices.add(i);
                    inPeriods = true;
                }
            } else {
                if (inPeriods) {
                    endIndices.add(i-1);
                    inPeriods = false;
                }
            }
        }
        if (inPeriods) {
            endIndices.add(n-1);
        }
        int f = 0;
        Iterator<Integer> beginIterator = beginIndices.iterator();
        Iterator<Integer> endIterator = endIndices.iterator();
        while (beginIterator.hasNext()) {
            int begin = beginIterator.next();
            int end = endIterator.next();
            f += end - begin;
        }

    }

}