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 a {
    public static void main(String[] args) throws IOException {
        FastReader scan = new FastReader();
        //PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("taming.out")));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        Task solver = new Task();
        //int t = scan.nextInt();
        int t = 1;
        for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
        out.close();
    }
    static class Task {
        public void solve(int testNumber, FastReader sc, PrintWriter pw) {
        	int n = sc.nextInt();
        	int m = sc.nextInt();
        	int[] arr = new int[n+2];
        	String str = sc.nextLine();
        	for(int i=1;i<=n;i++){
        		arr[i]=str.charAt(i-1)=='.'?1:0;
        	}
        	int sum = 0;
        	int temp = 0;
        	for(int i=1;i<=n;i++){
        		if(arr[i]==1)temp++;
        		else {
        			sum+=Math.max(0, temp-1);
        			temp=0;
        		}
        	}
        	sum+=Math.max(0, temp-1);
        	for(int i=0;i<m;i++){
        		int q = sc.nextInt();
        		int a =sc.next().charAt(0)=='.'?1:0;
        		if(arr[q]==a){
        			pw.println(sum);
        			continue;
        		}
        		if(a==1){
        			arr[q]=a;
        			if(arr[q-1]==1)sum++;
        			if(arr[q+1]==1)sum++;
        		}
        		else{
        			arr[q]=a;
        			if(arr[q-1]==1)sum--;
        			if(arr[q+1]==1)sum--;
        		}
        		pw.println(sum);
        	}
        }

    }
    static class tup implements Comparable<tup>{
        int a, b,c;
        tup(){};
        tup(int a, int b, int c){
            this.a=a;
            this.b=b;
            this.c=c;
        }

        @Override
        public int compareTo( tup o2) {
            return o2.b!=b?Integer.compare(o2.b,b):Integer.compare(o2.a,a);
        }
    }
    static void shuffle(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }
    static void shuffle(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        public FastReader(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        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());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}