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.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

/**
 * Created by Snayde on 05.01.2015.
 */
public class Main {

    public static void solve() {
        int n = IO.readInt();
        int m = IO.readInt();

        int numberOfQueries = 0;
        String s = IO.readString();
        boolean[] isDot = new boolean[n];
        for (int index = 0; index < n; ++index) {
            isDot[index] = s.charAt(index) == '.';
            if (index > 0 && isDot[index] && isDot[index - 1]) {
                numberOfQueries++;
            }
        }

        while (m-- > 0) {
            int index = IO.readInt() - 1;
            boolean dot = IO.readChar() == '.';
            if (dot && !isDot[index]) {
                if (index < n - 1 && isDot[index + 1]) {
                    numberOfQueries++;
                }
                if (index > 0 && isDot[index - 1]) {
                    numberOfQueries++;
                }
            }
            else if (!dot && isDot[index]) {
                if (index < n - 1 && isDot[index + 1]) {
                    numberOfQueries--;
                }
                if (index > 0 && isDot[index - 1]) {
                    numberOfQueries--;
                }
            }
            isDot[index] = dot;
            IO.println(numberOfQueries);
        }
    }

    // Input/Output ================================

    public static class IO {

        private static Scanner scanner;
        private static PrintWriter writer;

        public static void setUp(InputStream in, OutputStream out) {
            scanner = new Scanner(new java.io.BufferedInputStream(in));
            writer = new PrintWriter(out);
        }

        public static boolean isEmpty() {
            return !scanner.hasNext();
        }

        public static boolean hasNextLine() {
            return scanner.hasNextLine();
        }

        public static String readLine() {
            String line;
            try {
                line = scanner.nextLine();
            } catch (Exception e) {
                line = null;
            }
            return line;
        }

        public static String readString() {
            return scanner.next();
        }

        public static int readInt() {
            return scanner.nextInt();
        }

        public static double readDouble() {
            return scanner.nextDouble();
        }

        public static float readFloat() {
            return scanner.nextFloat();
        }

        public static long readLong() {
            return scanner.nextLong();
        }

        public static char readChar() {
            return scanner.findInLine("[^\\s]").charAt(0);
        }

        public static void print(Object o) {
            writer.print(o);
            writer.flush();
        }

        public static void println(Object o) {
            writer.println(o);
            writer.flush();
        }
    }

    // Algorithms ================================

    private static class Algo {

        private static int binSearch(int[] array, int n, int lo, int hi) {
            if (lo > hi) {
                return lo;
            }

            int mid = lo + (hi - lo) / 2;
            int cmp = (int) Math.signum(n - array[mid]);
            if (cmp < 0) {
                return binSearch(array, n, lo, mid - 1);
            } else if (cmp == 0) {
                return mid;
            } else {
                return binSearch(array, n, mid + 1, hi);
            }
        }

        private static void permutations(int n) {
            int[] p = new int[n];
            for (int index = 0; index < n; ++index) {
                p[index] = index + 1;
            }

            while (true) {
                IO.println(Arrays.toString(p));

                int k = n - 1;
                while (k > 0) {
                    if (p[k] < p[k - 1]) {
                        k--;
                    }
                    else {
                        break;
                    }
                }
                if (k == 0) {
                    break;
                }
                int j = n - 1;
                while (p[j] < p[k - 1]) {
                    j--;
                }

                int t = p[j];
                p[j] = p[k - 1];
                p[k - 1] = t;

                int i = 0;
                while (k + i < n - 1 - i) {
                    t = p[k + i];
                    p[k + i] = p[n - i - 1];
                    p[n - i - 1] = t;
                    i++;
                }
            }
        }

        private static final boolean isPrime(int number) {
            if (number <= 1) {
                return false;
            }
            if (number <= 3) {
                return true;
            }
            if (number % 2 == 0 || number % 3 == 0) {
                return false;
            }
            int i = 5;
            while (i * i <= number) {
                if (number % i == 0 || number % (i + 2) == 0) {
                    return false;
                }
                i += 6;
            }
            return true;
        }

        private static boolean isPalindrome(int number) {
            int n = number;
            int reverse = 0;
            while (n > 0) {
                reverse *= 10;
                reverse += (n % 10);
                n /= 10;
            }
            return reverse == number;
        }
    }

    public static final void main(String[] args) {
        IO.setUp(System.in, System.out);
        /*
        try {
            IO.setUp(new FileInputStream("encode.in"), new FileOutputStream("encode.out"));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
            return;
        }
        */
        solve();
    }
}