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.*;
import java.util.*;

public class Main implements Runnable {
    static boolean use_n_tests = false;
    static int stack_size = 1 << 27;

    void solve(FastScanner in, PrintWriter out, int testNumber) {
        int n, m;
        n = in.nextInt();
        m = in.nextInt();
        char[] s = ("$" + in.next() + "$").toCharArray();
        int ones = 0;
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            if (s[i + 1] == '.') {
                if (i == 0 || s[i] != '.') {
                    cnt++;
                }
                ones++;
            }
        }
        for (int i = 0; i < m; i++) {
            int a = in.nextInt();
            char c = in.next().toCharArray()[0];
            int change = 0;
            if (c == '.' && Character.isAlphabetic(s[a])) {
                s[a] = '.';
                ones++;
                if (s[a - 1] == '.' && s[a + 1] == '.') {
                    cnt--; // merge
                } else if ((s[a - 1] == '$' || Character.isAlphabetic(s[a - 1])) && (s[a + 1] == '$' || Character.isAlphabetic(s[a + 1]))) {
                    cnt++; // create new
                }
            } else if (Character.isAlphabetic(c) && s[a] == '.'){
                s[a] = c;
                ones--;
                if (s[a - 1] == '.' && s[a + 1] == '.') {
                    cnt++; // merge
                } else if ((s[a - 1] == '$' || Character.isAlphabetic(s[a - 1])) && (s[a + 1] == '$' || Character.isAlphabetic(s[a + 1]))) {
                    cnt--; // create new
                }
            }
            out.println(ones - cnt);
        }
    }

    // ****************************** template code ***********

    class Coeff {
        long mod;
        long[][] C;
        long[] fact;
        boolean cycleWay = false;

        Coeff(int n, long mod) {
            this.mod = mod;
            fact = new long[n + 1];
            fact[0] = 1;
            for (int i = 1; i <= n; i++) {
                fact[i] = i;
                fact[i] %= mod;
                fact[i] *= fact[i - 1];
                fact[i] %= mod;
            }
        }

        Coeff(int n, int m, long mod) {
            // n > m
            cycleWay = true;
            this.mod = mod;
            C = new long[n + 1][m + 1];
            for (int i = 0; i <= n; i++) {
                for (int j = 0; j <= Math.min(i, m); j++) {
                    if (j == 0 || j == i) {
                        C[i][j] = 1;
                    } else {
                        C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
                        C[i][j] %= mod;
                    }
                }
            }
        }

        public long C(int n, int m) {
            if (cycleWay) {
                return C[n][m];
            }
            return fC(n, m);
        }

        private long fC(int n, int m) {
            return (fact[n] * inv(fact[n - m] * fact[m] % mod)) % mod;
        }

        private long inv(long r) {
            if (r == 1)
                return 1;
            return ((mod - mod / r) * inv(mod % r)) % mod;
        }
    }

    class Pair {
        int first;
        int second;

        public int getFirst() {
            return first;
        }

        public int getSecond() {
            return second;
        }
    }

    class MultisetTree<T> {
        int size = 0;

        TreeMap<T, Integer> mp = new TreeMap<>();

        void add(T x) {
            mp.merge(x, 1, Integer::sum);
            size++;
        }

        void remove(T x) {
            if (mp.containsKey(x)) {
                mp.merge(x, -1, Integer::sum);
                if (mp.get(x) == 0) {
                    mp.remove(x);
                }
                size--;
            }
        }

        T greatest() {
            return mp.lastKey();
        }

        int size() {
            return size;
        }

        int diffSize() {
            return mp.size();
        }
    }

    class Multiset<T> {
        int size = 0;

        Map<T, Integer> mp = new HashMap<>();

        void add(T x) {
            mp.merge(x, 1, Integer::sum);
            size++;
        }

        void remove(T x) {
            if (mp.containsKey(x)) {
                mp.merge(x, -1, Integer::sum);
                if (mp.get(x) == 0) {
                    mp.remove(x);
                }
                size--;
            }
        }

        int size() {
            return size;
        }

        int diffSize() {
            return mp.size();
        }
    }

    static class Range {
        int l, r;
        int id;

        public int getL() {
            return l;
        }

        public int getR() {
            return r;
        }

        public Range(int l, int r, int id) {
            this.l = l;
            this.r = r;
            this.id = id;
        }
    }

    static class Array {
        static Range[] readRanges(int n, FastScanner in) {
            Range[] result = new Range[n];
            for (int i = 0; i < n; i++) {
                result[i] = new Range(in.nextInt(), in.nextInt(), i);
            }
            return result;
        }

        static public Integer[] read(int n, FastScanner in) {
            Integer[] out = new Integer[n];
            for (int i = 0; i < out.length; i++) {
                out[i] = in.nextInt();
            }
            return out;
        }

        static public int[] readint(int n, FastScanner in) {
            int[] out = new int[n];
            for (int i = 0; i < out.length; i++) {
                out[i] = in.nextInt();
            }
            return out;
        }
    }

    class Graph {
        List<List<Integer>> create(int n) {
            List<List<Integer>> graph = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                graph.add(new ArrayList<>());
            }
            return graph;
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream io) {
            br = new BufferedReader(new InputStreamReader(io));
        }

        public String line() {
            String result = "";
            try {
                result = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return result;
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

    void run_t_tests() {
        int t = in.nextInt();
        int i = 0;
        while (t-- > 0) {
            solve(in, out, i++);
        }
    }

    void run_one() {
        solve(in, out, -1);
    }

    @Override
    public void run() {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        if (use_n_tests) {
            run_t_tests();
        } else {
            run_one();
        }
        out.close();
    }

    static FastScanner in;
    static PrintWriter out;

    public static void main(String[] args) throws InterruptedException {
        Thread thread = new Thread(null, new Main(), "", stack_size);
        thread.start();
        thread.join();
    }
}