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:
/*
 * Remember a 7.0 student can know more than a 10.0 student.
 * Grades don't determine intelligence, they test obedience. 
 * I Never Give Up.
 */
import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.text.*;

import static java.lang.System.out;
import static java.util.Arrays.*;
import static java.lang.Math.*;
public class ContestMain {
	private static Reader in=new Reader();
	private static StringBuilder ans=new StringBuilder();
	private static long MOD=1000000007;//10^9+7 
//	private static final int N=100000; //10^5
//	private static final int LIM=26;
//	private static final double PI=3.1415;
//	private static ArrayList<Integer> v[]=new ArrayList[N];
//	private static int color[]=new int[N]; //For Graph Coloring
//	private static boolean mark[]=new boolean[N];
//	private static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
//	private static void dfs(int node){mark[node]=true;for(int x:v[node]){if(!mark[x]){dfs(x,v);}}}
//	private static long powmod(long x,long n,long m){
//		if(n==0)return 1;
//		else if(n%2==0)return(powmod((x*x)%m,n/2,m));
//		else return (x*(powmod((x*x)%m,(n-1)/2,m)))%m;
//	}
//	private static void shuffle(long [] arr) {
//		for (int i = arr.length - 1; i >= 2; i--) {
//			int x = new Random().nextInt(i - 1);
//	        long temp = arr[x];
//	        arr[x] = arr[i];
//	        arr[i] = temp;
//	    }
//	 }
//	private static void swap(int i,int j,int ar[]){
//		int temp=ar[i];
//		ar[i]=ar[j];
//		ar[j]=temp;
//	}
//	private static long gcd(long a, long b){
//		if(b==0)return a;
//		return gcd(b,a%b);
//	}
//	private static boolean check(int x,int y){
//		if((x>=0&&x<n)&&(y>=0&&y<m)&&mat[x][y]!='X'&&!visited[x][y])return true;
//		return false;
//	}
	public static void main(String[] args) throws IOException{
		int n=in.nextInt();
		int m=in.nextInt();
		char ar[]=in.next().toCharArray();
		int cnt=0,total=0;
		for(int i=n-1;i>=0;i--){
			if(ar[i]=='.')cnt++;
			else{
				total+=max(0,cnt-1);
				cnt=0;
			}
		}
		if(cnt!=0)total+=cnt-1;
		int pos;
		char c;
		while(m-->0){
			pos=in.nextInt()-1;
			c=in.next().charAt(0);
			if(c!='.'){
				if(ar[pos]!='.')ans.append(total+"\n");
				else{
					if(pos-1>=0&&ar[pos-1]=='.')total--;
					if(pos+1<n&&ar[pos+1]=='.')total--;
					ans.append(total+"\n");
				}
			}
			else{
				if(ar[pos]=='.')ans.append(total+"\n");
				else{
					if(pos-1>=0&&ar[pos-1]=='.')total++;
					if(pos+1<n&&ar[pos+1]=='.')total++;
					ans.append(total+"\n");
				}
			}
			ar[pos]=c;
		}
		out.println(ans);
	}
	static class Pair<T> implements Comparable<Pair>{
		int l;
		int r;
		int c;
		Pair(){
			l=0;
			r=0;
		}
		Pair(int k,int v,int p){
			l=k;
			r=v;
			c=p;
		}
		@Override
		public int compareTo(Pair o) {
			if(o.l!=l)return (int) (l-o.l);
			else return (r-o.r);
		}
		
	}
	static class Reader{
        BufferedReader br;
        StringTokenizer st;
        public Reader()
        {
            br = new BufferedReader(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());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}