Problem:
Score : 300 points

Problem Statement

You are given a string S of length N consisting of A, C, G and T. Answer the following Q queries:

Query i (1 \leq i \leq Q): You will be given integers l_i and r_i (1 \leq l_i < r_i \leq N). Consider the substring of S starting at index l_i and ending at index r_i (both inclusive). In this string, how many times does AC occurs as a substring?

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

2 \leq N \leq 10^5

1 \leq Q \leq 10^5

S is a string of length N.

Each character in S is A, C, G or T.

1 \leq l_i < r_i \leq N

Input

Input is given from Standard Input in the following format:

N Q
S
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

8 3
ACACTACG
3 7
2 3
1 8

Sample Output 1

2
0
3

Query 1: the substring of S starting at index 3 and ending at index 7 is ACTAC. In this string, AC occurs twice as a substring.

Query 2: the substring of S starting at index 2 and ending at index 3 is CA. In this string, AC occurs zero times as a substring.

Query 3: the substring of S starting at index 1 and ending at index 8 is ACACTACG. In this string, AC occurs three times as a substring.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
public class Main{
    public static void main(String[] args) throws IOException{
        Reader rd = new Reader();        
        int[] a = rd.ii();
      	int N = a[0],
      		Q = a[1];
      	String S = rd.s();
      	int[] l = new int[Q],
      	      r = new int[Q];
      	for(int i = 0; i < Q; i++){
        	a = rd.ii();
          	l[i] = a[0];
          	r[i] = a[1];
        }
      	char[] letter = S.toCharArray();
        int[] ac = new int[N+1];
        ac[0] = 0;
        for(int i = 1; i < N; i++){
            ac[i] = ac[i-1];
            if(letter[i-1] == 'A' && letter[i] == 'C') ac[i]++;
        }
      	for(int i = 0; i < Q; i++){
            print(ac[r[i]-1] - ac[l[i]-1]);
        }
    }
    static void print(Object o){
        System.out.println(o.toString());
    }
    static int max(int[] ii){
        int max = ii[0];
        for(int i : ii) if(max < i) max = i;
        return max;
    }
    static int min(int[] ii){
        int min = ii[0];
        for(int i : ii) if(i < min) min = i;
        return min;
    }
    static long max(long[] ll){
        long max = ll[0];
        for(long l : ll) if(max < l) max = l;     
        return max;
    }
    static long min(long[] ll){
        long min = ll[0];
        for(long l : ll) if(l < min) min = l;
        return min;
    }
    static double max(double[] dd){
        double max = dd[0];
        for(double d : dd) if(max < d) max = d;
        return max;
    }
    static double min(double[] dd){
        double min = dd[0];
        for(double d : dd) if(d < min) min = d;
        return min;
    }
    static class Reader extends BufferedReader{
        Reader(){
            super(new InputStreamReader(System.in));
        }
        String s() throws IOException{
            return readLine();
        }
        String[] ss() throws IOException{
            return readLine().split(" ");
        }
        int i() throws IOException{
            return Integer.parseInt(readLine());
        }
        int[] ii() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            int[] ii = new int[size];
            for(int j = 0; j < size; j++) ii[j] = Integer.parseInt(ss[j]);
            return ii;
        }
        int[] ii(int N) throws IOException{
            int[] ii = new int[N];
            for(int j = 0; j < N; j++) ii[j] = i();
            return ii;
        }
        long l() throws IOException{
            return Long.parseLong(readLine());
        }
        long[] ll() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            long[] ll = new long[size];
            for(int j = 0; j < size; j++) ll[j] = Long.parseLong(ss[j]);            
            return ll;
        }
        long[] ll(int N) throws IOException{
            long[] ll = new long[N];
            for(int j = 0; j < N; j++) ll[j] = i();            
            return ll;
        }
    }
}