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.io.*;
import java.math.*;
 public class a{
	public static void main( String [] args) throws IOException{
		FastScanner sc=new FastScanner();
		int n=sc.nextInt();
		int m=sc.nextInt();
		String s=" "+sc.next()+" ";
		int cnt=0;
		for(int i=0; i<s.length();i++)
			if(s.charAt(i)=='.' && s.charAt(i-1)=='.')cnt++;
		char [] ch=s.toCharArray();

		StringBuilder sb=new StringBuilder("");
	    for(int i=0;i<m;i++){
			int p=sc.nextInt();
			char c=sc.next().charAt(0);

			if(ch[p]=='.' && c!='.')
				cnt-=((ch[p-1]=='.'?1:0)+(ch[p+1]=='.'?1:0));
			
			if(ch[p]!='.' && c=='.')
				cnt+=((ch[p-1]=='.'?1:0)+(ch[p+1]=='.'?1:0));
			
			ch[p]=c;

			sb.append(cnt+"\n");
		}
		System.out.println(sb.toString().trim());
		
	}
}








































class FastScanner{	
	 private int BUFFER_SIZE = 1 << 16;
	
	private  DataInputStream din;
	
	private  byte[] buffer;
	
	private int bufferPointer, bytesRead;
	
	public FastScanner() {
		din = new DataInputStream(System.in);
		buffer = new byte[BUFFER_SIZE];
		bufferPointer = bytesRead = 0;
	}

	public FastScanner( String file_name) throws IOException {
		din = new DataInputStream(new FileInputStream(file_name));
		buffer = new byte[BUFFER_SIZE];
		bufferPointer = bytesRead = 0;
    }
    
	public String readLine() throws IOException {
		 byte[] buf = new byte[64];
		int cnt = 0, c;
		while ((c = read()) != -1) {
			if (c == '\n')
				break;
			buf[cnt++] = (byte) c;
		}
		return new String(buf, 0, cnt);
	}
    
    public String next() throws IOException{

		byte c = read();
		while(Character.isWhitespace(c)){
			c = read();
		}
		
		 StringBuilder builder = new StringBuilder();
		builder.append((char)c);
		c = read();
		while(!Character.isWhitespace(c)){
			builder.append((char)c);
			c = read();
		}
		
		return builder.toString();
	}

	public int nextInt() throws IOException {
		int ret = 0;
		byte c = read();
		while (c <= ' ')
			c = read();
		 boolean neg = (c == '-');
		if (neg)
			c = read();
		do {
			ret = ret * 10 + c - '0';
		} while ((c = read()) >= '0' && c <= '9');

		if (neg)
			return -ret;
		return ret;
	}
	
	public int[] nextIntArray( int n) throws IOException {
		 int arr[] = new int[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextInt();
		}
		return arr;
	}

	public long nextLong() throws IOException {
		long ret = 0;
		byte c = read();
		while (c <= ' ')
			c = read();
		 boolean neg = (c == '-');
		if (neg)
			c = read();
		do {
			ret = ret * 10 + c - '0';
		} while ((c = read()) >= '0' && c <= '9');
		if (neg)
			return -ret;
		return ret;
	}
	
	public long[] nextLongArray( int n) throws IOException {
		 long arr[] = new long[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextLong();
		}
		return arr;
	}

	public char nextChar() throws IOException{
		byte c = read();
		while(Character.isWhitespace(c)){
			c = read();
		}
		return (char) c;	
	}
	
	public double nextDouble() throws IOException {
		double ret = 0, div = 1;
		byte c = read();
		while (c <= ' ')
			c = read();
		 boolean neg = (c == '-');
		if (neg)
			c = read();

		do {
			ret = ret * 10 + c - '0';
		} while ((c = read()) >= '0' && c <= '9');

		if (c == '.') {
			while ((c = read()) >= '0' && c <= '9') {
				ret += (c - '0') / (div *= 10);
			}
		}

		if (neg)
			return -ret;
		return ret;
    }
    
	public double[] nextDoubleArray( int n) throws IOException {
		 double arr[] = new double[n];
		for(int i = 0; i < n; i++){
			arr[i] = nextDouble();
		}
		return arr;
	}

	private void fillBuffer() throws IOException {
		bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
		if (bytesRead == -1)
			buffer[0] = -1;
	}

	private byte read() throws IOException {
		if (bufferPointer == bytesRead)
			fillBuffer();
		return buffer[bufferPointer++];
	}

	public void close() throws IOException {
		if (din == null)
			return;
		din.close();
	}

}