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.InputStreamReader;
import java.io.IOException;
import java.util.Arrays;
import java.util.StringTokenizer;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.floor;
import static java.lang.Math.ceil;
import static java.lang.Math.pow;
import static java.lang.Math.log;
import static java.lang.Math.sqrt;

/* ------------ Theorems and Lemmas ------------ */

// Fermat's Little Theorem
// (1) ((a ** p) - a) % p = 0; if 'p' is prime
// (2) Further, if 'a % p != 0', then ((a ** (p - 1)) - 1) % p = 0

/* -------------- End of theorems -------------- */

public class Solution {

	// Digit DP
	private static int digit;
	private static int times;
	private static StringBuilder number;
	private static int[][][] digitDP;
	// End

	private static final String[] FACTS = new String[] { "", "", "2", "3", "322", "5", "53", "7", "7222", "7332" };

	public static void main(String[] args) {
		FastReader reader = new FastReader();
		int n = reader.nextInt();
		int m = reader.nextInt();
		char[] line = reader.next().toCharArray();
		int grp = 0;
		int cnt = 0;
		boolean[] point = new boolean[n + 2];
		for (int i = 0; i < n; ++i)
			if (line[i] == '.') {
				++cnt;
				if (i == 0 || line[i - 1] != '.')
					++grp;
				point[i + 1] = true;
			}
		StringBuilder sb = new StringBuilder("");
		while (m-- > 0) {
			int i = reader.nextInt();
			char c = reader.next().charAt(0);
			boolean a = point[i];
			boolean b = c == '.';
			if (a != b) {
				cnt += a ? -1 : 1;
				if (point[i - 1] && point[i + 1])
					grp += b ? -1 : 1;
				if (!point[i - 1] && !point[i + 1])
					grp += b ? 1 : -1;
			}
			point[i] = b;
			sb.append(cnt - grp + "\n");
		}
		reader.close();
		System.out.println(sb.toString().trim());
	}

	public static int ddp(int n) {
		number = new StringBuilder(String.valueOf(n));
		digitDP = new int[12][12][2];
		for (int[][] a : digitDP)
			for (int[] b : a)
				Arrays.fill(b, -1);
		return call(0, 0, 0);
	}

	// Digit DP
	public static int call(int pos, int cnt, int f) {
		if (cnt > times)
			return 0;
		if (pos == number.length())
			return cnt == times ? 1 : 0;
		if (digitDP[pos][cnt][f] != -1)
			return digitDP[pos][cnt][f];
		int res = 0;
		int lmt = f == 0 ? number.charAt(pos) - '0' : 9;
		for (int dgt = 0; dgt <= lmt; ++dgt)
			if ((dgt == digit ? 1 + cnt : cnt) <= times)
				res += call(1 + pos, dgt == digit ? 1 + cnt : cnt, f == 0 && dgt < lmt ? 1 : f);
		digitDP[pos][cnt][f] = res;
		return res;
	}

	// Binary Exponentiation
	public static int binpow(int a, int n, int mod) {
		// Iterative approach
		a %= mod;
		int result = 1;
		while (n > 0) {
			if ((n & 1) != 0)
				result = result * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return result;

		// Recursive approach
		/*
		 * if (n == 0l) return 1l; long res = binpow(a, n / 2, mod); if ((n & 1l) != 0l)
		 * return ((res % mod) * (res % mod) * (a % mod)) % mod; else return ((res %
		 * mod) * (res % mod)) % mod;
		 */
	}

	// GCD
	public static int gcd(int a, int b) {
		while (b > 0) {
			a %= b;
			int t = a;
			a = b;
			b = t;
		}
		return a;
	}

	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().trim());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] readArray(int size) {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = nextInt();
			return ar;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}
}