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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i, n) for (int i = 0; i < n; i++)
vector<ll> a;
// index が条件を満たすかどうか
bool isOK(int index, int key)
{
    if (a[index] >= key)
        return true;
    else
        return false;
}

// 汎用的な二分探索のテンプレ
int binary_search(int key)
{
    int ng = -1;            //「index = 0」が条件を満たすこともあるので、初期値は -1
    int ok = (int)a.size(); // 「index = a.size()-1」が条件を満たさないこともあるので、初期値は a.size()

    /* ok と ng のどちらが大きいかわからないことを考慮 */
    while (abs(ok - ng) > 1)
    {
        int mid = (ok + ng) / 2;

        if (isOK(mid, key))
            ok = mid;
        else
            ng = mid;
    }
    return ok;
}
int main()
{
    ll n, q;
    cin >> n >> q;
    string s;
    cin >> s;
    rep(i, n - 1)
    {
        if (s[i] == 'A' && s[i + 1] == 'C')
        {
            //この先頭のindexをどこかに保管する。
            //ループする時にrより大きくてlより小さいindexは何個あるのかをみる。
            a.push_back(i); //これは先頭なのでi+1も含むのが普通
        }
    }
    for (int i = 0; i < q; i++)
    {
        ll l, r;
        cin >> l >> r;
        l--;
        r--;
        //愚直にループを回すと10^5*10^5 前処理をする
        ll w = binary_search(l);     //これだと
        ll v = binary_search(r + 1); //
                                     //v-１のインデックスの数字+1とrの大小関係
                                     /*  if (a[v - 1] + 1 > r)
        {
            cout << v - w - 1 << endl;
        }*/
                                     /* else
        {
            cout << v - w << endl;
        }*/
        cout << v - w << endl;
    }
}