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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;


public class C {

	//IO
	static BufferedReader f;
	static PrintWriter out;
	static StringTokenizer st;
	
	final public static void main( String[] args ) throws IOException {
		f = new BufferedReader( new InputStreamReader( System.in ) );
		out = new PrintWriter( System.out );
		solve();
		f.close();
		out.close();
		System.exit( 0 );
	}
	
	final public static void solve() throws IOException {
		int strLen = nextInt();
		int numUpdates = nextInt();
		String str = nextToken();
		char[] array = str.toCharArray();
		
		int count = 0;
		for ( int i=1 ; i<str.length() ; ++i ) {
			if ( str.charAt( i ) == str.charAt( i-1 ) && str.charAt( i ) == '.' ) {
				++count;
			}
		}
		
		for ( int i=0 ; i<numUpdates ; ++i ) {
			int idx = nextInt();
			--idx;
			String letter = nextToken();
			if ( letter.equals( "." ) ) {
				if ( array[ idx ] != '.' ) {
					if ( idx > 0 ) {
						if ( array[ idx-1 ] == '.' ) {
							++count;
						}
					}
					if ( idx < strLen-1 ) {
						if ( array[ idx+1 ] == '.' ) {
							++count;
						}
					}
				}
			}
			else {
				if ( array[ idx ] == '.' ) {
					if ( idx > 0 ) {
						if ( array[ idx-1 ] == '.' ) {
							--count;
						}
					}
					if ( idx < strLen-1 ) {
						if ( array[ idx+1 ] == '.' ) {
							--count;
						}
					}
				}
			}
			array[ idx ] = letter.charAt( 0 );
			System.out.println( count );
		}
		
	}
	
	final public static String nextToken() throws IOException {
		while ( st == null || !st.hasMoreTokens() ) {
			st = new StringTokenizer( f.readLine() );
		}
		return st.nextToken();
	}
	
	final public static int nextInt() throws IOException {
		return Integer.parseInt( nextToken() );
	}
	
	final public static long nextLong() throws IOException {
		return Long.parseLong( nextToken() );
	}
	
	final public static double nextDouble() throws IOException {
		return Double.parseDouble( nextToken() );
	}
	
	final public static boolean nextBoolean() throws IOException {
		return Boolean.parseBoolean( nextToken() );
	}
}