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.text.DecimalFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
import java.io.*;

public class abc {
	static PrintWriter pw;

	/*
	 * static long inv[]=new long[1000001]; static long dp[]=new long[1000001];
	 */
	/// MAIN FUNCTION///
	public static void main(String[] args) throws Exception {

		FastReader sc = new FastReader();
		pw = new PrintWriter(System.out);
		// use arraylist as it use the concept of dynamic table(amortized analysis)
		// Arrays.stream(array).forEach(a -> Arrays.fill(a, 0));
		/* List<Integer> l1 = new ArrayList<Integer>(); */
		//Random rand = new Random();
		int tst = 1;
		//pw.println(tst);
		
		while (tst-- > 0) {
			//String s=sc.nextLine()
			int n=sc.nextInt();
			int m=sc.nextInt();
			String s=sc.nextLine();
			char ch[]=s.toCharArray();
			int ans=0;
			int count=0;
			for(int i=0;i<s.length();i++)
			{
				if(ch[i]=='.')
					count++;
				else
				{
					if(count>1)
						ans+=(count-1);
					count=0;
				}
			}
			if(count>1)
			ans+=count-1;
			int res=0;
			for(int i=0;i<m;i++)
			{
				res=0;
				int x=sc.nextInt();
				char c=sc.next().charAt(0);
				if(n==1)
					res=0;
				else if(x==1)
				{
					if(c!='.' && ch[x-1]=='.')
					{
						if(ch[x]=='.')
							res-=1;
					}
					if(c=='.' && ch[x-1]!='.')
					{
						if(ch[x]=='.')
							res+=1;
					}
				}
				else if(x==s.length())
				{
					if(c!='.' && ch[x-1]=='.')
					{
						if(ch[x-2]=='.')
							res-=1;
					}
					if(c=='.' && ch[x-1]!='.')
					{
						if(ch[x-2]=='.')
							res+=1;
					}
				}
				else
				{
					if(c!='.' && ch[x-1]=='.')
					{
						if(ch[x-2]=='.' && ch[x]=='.')
							res-=2;
						else if(ch[x-2]=='.' || ch[x]=='.')
							res-=1;
					}
					if(c=='.' && ch[x-1]!='.')
					{
						if(ch[x-2]=='.' && ch[x]=='.')
							res+=2;
						else if(ch[x-2]=='.' || ch[x]=='.')
							res+=1;
					}
				}
				ch[x-1]=c;
				ans+=res;
				pw.println(ans);
			}
		}
		pw.flush();
	}

	public static int helper(int b[][], int k, int m, int n) {
		int res = 0, f = 0;
		for (int i = 0; i < m; i++) {
			for (int j = 0; j < n; j++) {
				if (k == b[i][j]) {
					res = j;
					f = -1;
					break;
				}
			}
			if (f == -1)
				break;
		}
		return res;
	}

	public static void debugger() {
		Random rand = new Random();
		int tst = (int) (Math.abs(rand.nextInt()) % 2 + 1);
		pw.println(tst);
		while (tst-- > 0) {
			int n = (int) (Math.abs(rand.nextInt()) % 5 + 1);
			pw.println(n);
			for (int i = 0; i < n; i++) {
				pw.print((int) (Math.abs(rand.nextInt()) % 6 + 1) + " ");
			}
			pw.println();
		}
	}

	static void recursion(int n) {
		if (n == 1) {
			pw.print(n + " ");
			return;
		}
		// pw.print(n+" "); gives us n to 1
		recursion(n - 1);
		// pw.print(n+" "); gives us 1 to n
	}

	// ch.charAt(i)+"" converts into a char sequence
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			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;
		}

		/* CREATED BY ME */
		int[] readArray(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = nextInt();
			return a;
		}

		long[] readLongArray(int n) {
			long[] a = new long[n];
			for (int i = 0; i < n; i++)
				a[i] = nextLong();
			return a;
		}
	}

	public static long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	public static boolean isPrime(long n) {
		if (n == 2)
			return true;
		long i = 2;
		while (i * i <= n) {
			if (n % i == 0)
				return false;
			i++;
		}
		return true;
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static double max(double x, double y) {
		return Math.max(x, y);
	}

	static int min(int x, int y) {
		return Math.min(x, y);
	}

	static int abs(int x) {
		return Math.abs(x);
	}

	static long abs(long x) {
		return Math.abs(x);
	}

	static int log2(int N) {
		int result = (int) (Math.log(N) / Math.log(2));
		return result;
	}

	static long max(long x, long y) {
		return Math.max(x, y);
	}

	static long min(long x, long y) {
		return Math.min(x, y);
	}

	public static class pair {
		long x;
		long y;

		public pair(long a, long b) {
			x = a;
			y = b;
		}
	}

	public static class Comp implements Comparator<pair> {
		public int compare(pair a, pair b) {
			long ans = a.x - b.x;
			if (ans > 0)
				return 1;
			if (ans < 0)
				return -1;
			return 0;
		}
	}

	// modular exponentiation
	public static long fastExpo(long a, int n, int mod) {
		if (n == 0)
			return 1;
		else {
			if ((n & 1) == 1) {
				long x = fastExpo(a, n / 2, mod);
				return (((a * x) % mod) * x) % mod;
			} else {
				long x = fastExpo(a, n / 2, mod);
				return (((x % mod) * (x % mod)) % mod) % mod;
			}
		}
	}

	public static long modInverse(long n, int p) {
		return fastExpo(n, p - 2, p);
	}

	/*
	 * public static void extract(ArrayList<Integer> ar, int k, int d) { int c = 0;
	 * for (int i = 1; i < k; i++) { int x = 0; boolean dm = false; while (x > 0) {
	 * long dig = x % 10; x = x / 10; if (dig == d) { dm = true; break; } } if (dm)
	 * ar.add(i); } }
	 */

	public static int[] prefixfuntion(String s) {
		int n = s.length();
		int z[] = new int[n];
		for (int i = 1; i < n; i++) {
			int j = z[i - 1];
			while (j > 0 && s.charAt(i) != s.charAt(j))
				j = z[j - 1];
			if (s.charAt(i) == s.charAt(j))
				j++;
			z[i] = j;
		}
		return z;
	}

	public static void dfs(int index, boolean vis[], int a[], int b[], int n) {
		vis[index] = true;
		for (int i = 0; i < n; i++) {
			if (!vis[i] && (a[i] == a[index] || b[i] == b[index]))
				dfs(i, vis, a, b, n);
		}
	}

// counts the set(1) bit of a number
	public static long countSetBitsUtil(long x) {
		if (x <= 0)
			return 0;
		return (x % 2 == 0 ? 0 : 1) + countSetBitsUtil(x / 2);
	}

//tells whether a particular index has which bit of a number	
	public static int getIthBitsUtil(int x, int y) {
		return (x & (1 << y)) != 0 ? 1 : 0;
	}

	public static void swap(long x, long y) {
		x = x ^ y;
		y = y ^ x;
		x = x ^ y;
	}

	/*
	 * static long mod_inverse(long a, long b) { long mod=(long) (1000000007); long
	 * ans=1; while(b>0) { if((b%2)==0) ans=(ans*a) %mod; b/=2; a=(a*a) % mod;
	 * 
	 * } return ans; }
	 */

	public static double decimalPlaces(double sum) {
		DecimalFormat df = new DecimalFormat("#.00");
		String angleFormated = df.format(sum);
		double fin = Double.parseDouble(angleFormated);
		return fin;
	}

//use collections.swap for swapping 
	static boolean isSubSequence(String str1, String str2, int m, int n) {
		int j = 0;

		for (int i = 0; i < n && j < m; i++)
			if (str1.charAt(j) == str2.charAt(i))
				j++;

		return (j == m);
	}

	static long sum(long n) {
		long s2 = 0, max = -1, min = 10;
		while (n > 0) {
			s2 = (n % 10);
			min = min(s2, min);
			max = max(s2, max);
			n = n / 10;
		}
		return max * min;
	}

	static long pow(long base, long power) {
		if (power == 0) {
			return 1;
		}
		long result = pow(base, power / 2);
		if (power % 2 == 1) {
			return result * result * base;
		}
		return result * result;
	}

	// return the hash value of a string
	static long compute_hash(String s) {
		long val = 0;
		long p = 31;
		long mod = (long) (1000000007);
		long pow = 1;
		for (int i = 0; i < s.length(); i++) {
			char ch = s.charAt(i);
			val = (val + (int) (ch - 'a' + 1) * pow) % mod;
			pow = (pow * p) % mod;
		}
		return val;
	}

	// return a prefix sum array of hash values
	/*
	 * static void prefix_array_value(String s) { long p=31; long mod=(long)
	 * (1000000007); long pow=1; inv[0]=1; dp[0]=(long)(s.charAt(0)-'a'+1);
	 * 
	 * for(int i=1;i<s.length();i++) { char ch=s.charAt(i); pow= ( pow*p ) % mod;
	 * inv[i]=mod_inverse(pow, mod-2); dp[i]=(dp[i-1]+ (long)(ch- 'a' +1)*pow)%mod;
	 * } }
	 */

	/*
	 * static long substringHash(int L , int R ) { long result = dp[R]; long
	 * mod=(long) (1000000007); if(L > 0) result = (result - dp[L-1] + mod) % mod;
	 * result= (result * inv[L]) % mod; return result; }
	 */
}