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.*;

public class Main {

    static MyScanner in;
    static PrintWriter out;
    //static Timer t = new Timer();

    public static void main(String[] args) throws IOException {
        in = new MyScanner();
        out = new PrintWriter(System.out, true);

        int n = in.nextInt(), m = in.nextInt();
        char[] s = in.next().toCharArray();

        int cnt = 0;
        for (int i = 1; i < n; i++)
            if (s[i] == '.' && s[i - 1] == '.')
                cnt++;

        while (m-- != 0) {
            int i = in.nextInt() - 1;
            char ch = in.next().charAt(0);

            if (ch != '.' && s[i] != '.' || ch == '.' && s[i] == '.') {
                out.println(cnt);
                continue;
            }

            int del = ch == '.' ? 1 : -1;
            if (i > 0 && s[i - 1] == '.')
                cnt += del;
            if (i < n - 1 && s[i + 1] == '.')
                cnt += del;

            s[i] = ch;
            out.println(cnt);
        }
    }

}

class MyScanner {

    private final BufferedReader br;
    private StringTokenizer st;

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

    public MyScanner(String path) throws IOException {
        br = new BufferedReader(new FileReader(new File(path)));
        //br = new BufferedReader(new InputStreamReader(new FileInputStream(
        //new File(path)), "cp1251"));
    }

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

    String nextLine() throws IOException {
        st = null;
        return br.readLine();
    }

    boolean hasNext() {
        if (st != null && st.hasMoreElements())
            return true;

        try {
            while (st == null || !st.hasMoreElements())
                st = new StringTokenizer(br.readLine());
        } catch (Exception e) {
            return false;
        }

        return true;
    }

    String[] nextStrings(int n) throws IOException {
        String[] arr = new String[n];
        for (int i = 0; i < n; i++)
            arr[i] = next();
        return arr;
    }

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

    int[] nextInts(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    Integer[] nextIntegers(int n) throws IOException {
        Integer[] arr = new Integer[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    int[][] next2Ints(int n, int m) throws IOException {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextInt();
        return arr;
    }

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

    long[] nextLongs(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLong();
        return arr;
    }

    long[][] next2Longs(int n, int m) throws IOException {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextLong();
        return arr;
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next().replace(',', '.'));
    }

    double[] nextDoubles(int size) throws IOException {
        double[] arr = new double[size];
        for (int i = 0; i < size; i++)
            arr[i] = nextDouble();
        return arr;
    }

    boolean nextBool() throws IOException {
        String s = next();
        if (s.equalsIgnoreCase("true") || s.equals("1"))
            return true;

        if (s.equalsIgnoreCase("false") || s.equals("0"))
            return false;

        throw new IOException("Boolean expected, String found!");
    }
}