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.util.*;
    import java.lang.Math; 
    import java.util.Arrays;
    import java.util.Scanner;
    import java.util.stream.IntStream;
    import java.io.OutputStream;
    import java.io.IOException;
    import java.io.InputStream;
    import java.io.PrintWriter;
    import java.util.Arrays;
     
    public class Main {
         static int oo = (int)1e9;
      public static void main(String[] args) {
    	       InputStream inputStream = System.in;
            OutputStream outputStream = System.out;
            Scanner in = new Scanner(inputStream);
            PrintWriter out = new PrintWriter(outputStream);
            TaskA solver = new TaskA();
            solver.solve(1, in, out);
            out.close();
    	}
    	
    	 static class TaskA {
       public void solve(int testNumber, Scanner in, PrintWriter out) {
        //  int t = in.nextInt();
       //     while(t-->0){
           int n = in.nextInt(),m= in.nextInt();
           char[] s = new char[n + 2];
           System.arraycopy(in.next().toCharArray(), 0, s, 1, n);
           int cc=0,dots=0;
           for(int i=1;i<=n;i++){
                if(s[i]=='.'){
                    dots++;
                     if(s[i-1]!='.'){
                          cc++;
                     }
                }else{
                     s[i]=0;
                }
           }
                
           //   out.println(dots+" "+cc);
           for(int i=0;i<m;i++){
                int x = in.nextInt();
                char what = in.next().charAt(0);
                 
                if(what!='.'){
                     what=0;
                }
                if(what!=s[x]){
                     s[x]=what;
                      if(what=='.'){
                           dots++;
                           if(s[x-1]==0 && s[x+1]==0){
                                cc++;
                           }
                           if(s[x-1]=='.'&&s[x+1]=='.'){
                                cc--;
                           }
                      }else{
                           dots--;
                           if(s[x-1]==0 && s[x+1]==0){
                                cc--;
                           }
                           if(s[x-1]=='.' && s[x+1]=='.'){
                                cc++;
                           }
                      }
                  }
                  out.println(dots-cc);
            }
          
         // }      
       }
    	}     
    	  	
    public int factorial(int n) {
            int fact = 1;
            int i = 1;
           while(i <= n) {
             fact *= i;
             i++;
           }
            return fact;
    }
     
  
    public static long gcd(long x,long y)
    {
    	if(x%y==0)
    		return y;
    	else
    		return gcd(y,x%y);
    }
    public static int gcd(int x,int y)
    {
    	if(x%y==0)
    		return y;
    	else 
    		return gcd(y,x%y);
    }
    public static int abs(int a,int b)
    {
    	return (int)Math.abs(a-b);
    }
    public static long abs(long a,long b)
    {
    	return (long)Math.abs(a-b);
    }
    public static int max(int a,int b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static int min(int a,int b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
    public static long max(long a,long b)
    {
    	if(a>b)
    		return a;
    	else
    		return b;
    }
    public static long min(long a,long b)
    {
    	if(a>b)
    		return b;
    	else 
    		return a;
    }
     
     
    public static long pow(long n,long p,long m)
    {
    	 long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;
    	    if(result>=m)
    	    result%=m;
    	    p >>=1;
    	    n*=n;
    	    if(n>=m)
    	    n%=m;
    	}
    	return result;
    }
    public static long pow(long n,long p)
    {
    	long  result = 1;
    	  if(p==0)
    	    return 1;
    	if (p==1)
    	    return n;
    	while(p!=0)
    	{
    	    if(p%2==1)
    	        result *= n;	    
    	    p >>=1;
    	    n*=n;	    
    	}
    	return result;
     
    }
     
    static long sort(int a[]){  
         int n=a.length;
    	int b[]=new int[n];	
    	return mergeSort(a,b,0,n-1);
         
    }
     
    static long mergeSort(int a[],int b[],long left,long right){
         long c=0;
         if(left<right){   
          long mid=left+(right-left)/2;
    	 c= mergeSort(a,b,left,mid);
    	 c+=mergeSort(a,b,mid+1,right);
    	 c+=merge(a,b,left,mid+1,right); 
         }	
    	return c;	 
    }
    static long merge(int a[],int  b[],long left,long mid,long right){
         long c=0;int i=(int)left;int j=(int)mid; int k=(int)left;
         while(i<=(int)mid-1&&j<=(int)right){ 
              if(a[i]<=a[j]){
                   b[k++]=a[i++]; 
              }
              else{ 
                   b[k++]=a[j++];c+=mid-i;
              }
         }
         
         while (i <= (int)mid - 1)   
                 b[k++] = a[i++]; 
         while (j <= (int)right)
                 b[k++] = a[j++];
         for (i=(int)left; i <= (int)right; i++) 
    	        a[i] = b[i];
    	        return c; 
         }
     
  }