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:
//package round316;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;

public class C {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni(), q = ni();
		LST lst = new LST(n+2);
		lst.set(0);
		lst.set(n+1);
		char[] s = ns().toCharArray();
		int dot = 0;
		int ret = 0;
		for(int i = 0;i < n;i++){
			if(s[i] != '.'){
				lst.set(i+1);
				dot = 0;
			}else{
				dot++;
				if(dot >= 2)ret++;
			}
		}
		for(int z = 0;z < q;z++){
			int pos = ni();
			char c = nc();
			if(c != '.'){
				if(s[pos-1] == '.'){
					int bef = lst.prev(pos);
					int aft = lst.next(pos);
					ret -= Math.max(aft-bef-2, 0);
					ret += Math.max(aft-pos-2, 0);
					ret += Math.max(pos-bef-2, 0);
					lst.set(pos);
				}
			}else{
				if(s[pos-1] != '.'){
					lst.unset(pos);
					int bef = lst.prev(pos);
					int aft = lst.next(pos);
					ret += Math.max(aft-bef-2, 0);
					ret -= Math.max(aft-pos-2, 0);
					ret -= Math.max(pos-bef-2, 0);
				}
			}
			out.println(ret);
			
			s[pos-1] = c;
		}
	}
	
	public static class LST {
		public long[][] set;
		public int n;
//		public int size;
		
		public LST(int n) {
			this.n = n;
			int d = 1;
			for(int m = n;m > 1;m>>>=6, d++);
			
			set = new long[d][];
			for(int i = 0, m = n>>>6;i < d;i++, m>>>=6){
				set[i] = new long[m+1];
			}
//			size = 0;
		}
		
		// [0,r)
		public LST setRange(int r)
		{
			for(int i = 0;i < set.length;i++, r=r+63>>>6){
				for(int j = 0;j < r>>>6;j++){
					set[i][j] = -1L;
				}
				if((r&63) != 0)set[i][r>>>6] |= (1L<<r)-1;
			}
			return this;
		}
		
		// [0,r)
		public LST unsetRange(int r)
		{
			if(r >= 0){
				for(int i = 0;i < set.length;i++, r=r+63>>>6){
					for(int j = 0;j < r+63>>>6;j++){
						set[i][j] = 0;
					}
					if((r&63) != 0)set[i][r>>>6] &= ~((1L<<r)-1);
				}
			}
			return this;
		}
		
		public LST set(int pos)
		{
			if(pos >= 0 && pos < n){
//				if(!get(pos))size++;
				for(int i = 0;i < set.length;i++, pos>>>=6){
					set[i][pos>>>6] |= 1L<<pos;
				}
			}
			return this;
		}
		
		public LST unset(int pos)
		{
			if(pos >= 0 && pos < n){
//				if(get(pos))size--;
				for(int i = 0;i < set.length && (i == 0 || set[i-1][pos] == 0L);i++, pos>>>=6){
					set[i][pos>>>6] &= ~(1L<<pos);
				}
			}
			return this;
		}
		
		public boolean get(int pos)
		{
			return pos >= 0 && pos < n && set[0][pos>>>6]<<~pos<0;
		}
		
		public int prev(int pos)
		{
			for(int i = 0;i < set.length && pos >= 0;i++, pos>>>=6, pos--){
				int pre = prev(set[i][pos>>>6], pos&63);
				if(pre != -1){
					pos = pos>>>6<<6|pre;
					while(i > 0)pos = pos<<6|63-Long.numberOfLeadingZeros(set[--i][pos]);
					return pos;
				}
			}
			return -1;
		}
		
		public int next(int pos)
		{
			for(int i = 0;i < set.length && pos>>>6 < set[i].length;i++, pos>>>=6, pos++){
				int nex = next(set[i][pos>>>6], pos&63);
				if(nex != -1){
					pos = pos>>>6<<6|nex;
					while(i > 0)pos = pos<<6|Long.numberOfTrailingZeros(set[--i][pos]);
					return pos;
				}
			}
			return -1;
		}
		
		private static int prev(long set, int n)
		{
			long h = Long.highestOneBit(set<<~n);
			if(h == 0L)return -1;
			return Long.numberOfTrailingZeros(h)-(63-n);
		}
		
		private static int next(long set, int n)
		{
			long h = Long.lowestOneBit(set>>>n);
			if(h == 0L)return -1;
			return Long.numberOfTrailingZeros(h)+n;
		}
		
		@Override
		public String toString()
		{
			List<Integer> list = new ArrayList<Integer>();
			for(int pos = next(0);pos != -1;pos = next(pos+1)){
				list.add(pos);
			}
			return list.toString();
		}
	}

	
	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new C().run(); }
	
	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}