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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define repz(i, z, n) for (decltype(n) i = (z); i < (n); i++)
#define rep(i, n) for (decltype(n) i = 0; i < (n); i++)

#define REPZ(i, z, n) for (decltype(n) i = (z); i <= (n); i++)
#define REP(i, n) for (decltype(n) i = 0; i <= (n); i++)
#define ALL(v) v.begin(), v.end()
#define SIZE(v) ((ll)(v).size())
#define MAX(v) (*max_element(ALL(v)))
#define MIN(v) (*min_element(ALL(v)))
#define INF 1000000000000  // 1E+12

// 略
#define PB push_back
#define F first
#define S second
#define MP make_pair
#define READ cin >>
#define PRINT cout <<
#define PYES cout << "YES" << endl
#define pyes cout << "Yes" << endl
#define POK cout << "OK" << endl
#define PNO cout << "NO" << endl
#define pno cout << "No" << endl
#define PNG cout << "NG" << endl
#define ret return 0

using namespace std;

// デバッグ用
void debug_out() { cout << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cout << H << " ";
  debug_out(T...);
}

template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

#ifdef _DEBUG
#define debug(...) debug_out(__VA_ARGS__)
#else
#define debug(...)
#endif

// typedef
typedef long long ll;
typedef unsigned long long ul;
typedef long double ld;
typedef pair<ll, ll> pl;

// const
// 円周率
const ld pi = acos(-1);
// mod用
// 1E+9 +7
const ll mod = (int)1e+9 + 7;
// 1s間で可能なループ回数2E+8回
// int        最大値2147483647         〜2E+9
// long long  最大値9223372036854775807〜9E+18
// 総和accumulate
// 単純リストvector<type> push_back
// 連想配列map<key,value> insert,
// 集合set<type>
// pair キーとバリューを格納 firstとsecondでアクセス
// 小数点指定 cout << fixed << setprecision(15) << endl;
// to_string<->stoi
// stoll(longlong)
// stod(double)
// deep copy v1->v2 copy(v1.begin(), v1.end(), back_inserter(v2) );
// 大文字65-90(-32)
// 小文字97-122(+32)
//  priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>>>
//  ba;
signed main() {
  // faster
  // C言語流の入出力は使用できない
  ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  int q;
  cin >> n >> q;
  string s;
  cin >> s;
  vector<int> dp(SIZE(s), 0);
  bool flag = false;
  repz(i, 1, SIZE(s)) { dp[i] = dp[i - 1] + (s[i - 1] == 'A' and s[i] == 'C'); }
  rep(i, q) {
    int l, r;
    cin >> l >> r;
    l--;
    r--;
    cout << dp[r] - dp[l] << endl;
  }
  ret;
}