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 Replacement
{
    public static void main(String []av) {
        Scanner _s = new Scanner();
        int n = _s.nextInt();
        int m = _s.nextInt();
        char [] s = _s.next().toCharArray();
        TreeMap<Integer, Integer> dot = new TreeMap<Integer, Integer>();
        int currentDotStart = -1;
        int currentDotLength = 0;
        int f = 0;
        for (int i = 0; i < n; i++) {
            if (s[i] == '.') {
                if (currentDotStart == -1) {
                    currentDotStart = i;
                    currentDotLength = 1;
                } else {
                    currentDotLength++;
                }
            } else {
                if (currentDotStart != -1) {
                    dot.put(currentDotStart, currentDotLength);
                    f += currentDotLength - 1;
                }

                currentDotStart = -1;
            }
        }
        if (currentDotStart != -1) {
            dot.put(currentDotStart, currentDotLength);
            f += currentDotLength - 1;
        }

        //System.out.println("initial f: " + f);
        //System.out.println("initial dot: " + dot);
        StringBuilder out = new StringBuilder();
        for (int i = 0; i < m; i++) {
            int x = _s.nextInt() - 1;
            Map.Entry<Integer, Integer> e = dot.floorEntry(x);
            char c = _s.next().charAt(0);

            if (e != null) {
                int left = e.getKey();
                int right = left + e.getValue();
                if (c == '.') {
                    //System.out.println("next interval: " + left + " to " + right + " x at " + x);
                    if (x == right) {
                        dot.remove(e.getKey());
                        dot.put(left, x - left + 1);
                        //System.out.println("expand by 1 on right, dot now " + dot);
                        f++;
                        f += coalesce(dot, left);
                    }
                    if (x > right) {
                        dot.put(x, 1);
                        f += coalesce(dot, x);
                    }
                    // else it's already a dot
                } else {
                    // it's a letter, split the interval
                    if (left <= x && x <= right) {
                        int prevf = e.getValue() - 1;
                        int newf = 0;
                        dot.remove(e.getKey());
                        if (x-left > 0) {
                            dot.put(left, x-left);
                            newf += x-left-1;
                        }
                        if (right - (x+1) > 0) {
                            dot.put(x+1, right-(x+1));
                            newf += right-(x+1)-1;
                        }
                        f += (newf - prevf);
                    }
                }
            } else {
                //System.err.println("hit left at " + x);
                // at the left most
                if (c == '.') {
                    dot.put(x, 1);
                    f += coalesce(dot, x);
                }
            }
            //System.out.println("dot now: " + dot);
            out.append(Integer.toString(f));
            out.append("\n");
        }
        System.out.print(out.toString());
    }

    static int coalesce(TreeMap<Integer, Integer> dot, int p) {
        Map.Entry<Integer, Integer> e = dot.floorEntry(p);
        Map.Entry<Integer, Integer> n = dot.floorEntry(e.getKey() + e.getValue());
        if (p + e.getValue() == n.getKey()) {
            int oldf = e.getValue() - 1;
            oldf += n.getValue() - 1;
            int newLength = e.getValue() + n.getValue();
            int newf = newLength - 1;
            dot.remove(p);
            dot.remove(n.getKey());
            dot.put(p, newLength);
            // System.out.println("coalescing: new: " + newf + " -old " + oldf);
            return newf - oldf;
        }
        return 0;
    }

    //-----------Scanner class for faster input----------
    /* Provides similar API as java.util.Scanner but does not
     * use regular expression engine.  
     */
    public static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        public Scanner(Reader in) {
            br = new BufferedReader(in);
        }

        public Scanner() { this(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()); }

        // Slightly different from java.util.Scanner.nextLine(),
        // which returns any remaining characters in current line,
        // if any.  
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
    //--------------------------------------------------------    
}