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 C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define REP(i, n) for(int i = 1; i < (int)(n); i++)
typedef long long ll;
#define NP next_permutation
#define MP make_pair

int dp[100];

//最大公約数
int gcd(int x, int y){
  if(x % y == 0) return y;
  return gcd(y, x % y);
}
//最小公倍数
int lcm(int x, int y){
  return x * y / gcd(x, y);
}
//フィボナッチ数列
//配列dpを-1で初期化する
int fib(int n){
  if(n == 0 || n == 1) return dp[n] = 1;
  if(dp[n] != -1) return dp[n];
  return dp[n] = fib(n - 1) + fib(n - 2);
}
//階乗計算(factorial)
int fact(int n){
  if(n = 1) return 1;
  return n * fact(n - 1);
}
//累乗計算(Exponentiation)
//x**y
ll expo(int x, int y){
  ll ans = 1;
  rep(i, y){
    ans *= x;
  }
  return ans;
}
//桁数
int keta(ll x){
  if(x == 0) return 1;
  int res = 0;
  while(x){
    x /= 10;
    res++;
  }
  return res;
}


int main(){
  int n, q;
  cin >> n >> q;
  vector<char> s(n);
  rep(i,n) cin >> s[i];

  rep(i, q){
    int l, r;
    cin >> l >> r;
    int ans = 0;
    for(int i = l-1; i < r-1; i++){
      if(s[i] == 'A' && s[i+1] == 'C') ans ++;
    }
    cout << ans << endl;
  }
}