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: Runtime Error)

#include<algorithm>
#include<iomanip>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<vector>
#include<bitset>
#include<cmath>
#define FOR(i,a,b) for(int i=(a); i<int(b); ++i)
#define RFOR(i,a,b) for(int i=(b)-1; i>=int(a); --i)
#define rep(i,n) FOR(i,0,n)
#define rep1(i,n) FOR(i,1,int(n)+1)
#define rrep(i, n) RFOR(i, 0, n)
#define rrep1(i, n) RFOR(i, 1, int(n) + 1)
#define all(c) begin(c), end(c)
#define al(d) d.begin(),d.end()
#define fill(n) std::setfill('0')<<std::right<<std::setw(n)
template <typename T>
using vec = std::vector<T>;
using vi = vec<int>;
using vvi = vec<vi>;
using pii = std::pair<int, int>;
using ll = long long;
using ld = long double;
using namespace std;
const int MOD = 1000000007;

template <typename T>
void putv(vector<T>& V){
	// cout << "The elements in the vector are: " << endl;
	for(auto x: V)
		cout << x << " ";
	cout << endl;
}
template <class T>
vector<T> getv(int n){
	vector<T> vec;
	rep(i,n){
		T input;
		cin >> input;
		vec.emplace_back(input);
	}
	return vec;
}

using AdjList = vector<vector<int>>;
using Weight = int;
using Flow = int;
struct Edge {
	int src, dst;
	Weight weight;
	Flow cap;
	Edge() : src(0), dst(0), weight(0) {}
	Edge(int s, int d, Weight w) : src(s), dst(d), weight(w) {}
};
using Edges = std::vector<Edge>;
using Graph = std::vector<Edges>;
using Array = std::vector<Weight>;
using Matrix = std::vector<Array>;
void add_edge(Graph &g, int a, int b, Weight w = 1) {
	g[a].emplace_back(a, b, w);
	g[b].emplace_back(b, a, w);
}
void add_arc(Graph &g, int a, int b, Weight w = 1) { g[a].emplace_back(a, b, w); }
// Recieve edge(=(int,int)) as an adj list.
AdjList getadj(int N, int M){
	// Initialize all lists of the graph as a empty vector.
	vector<int> init;
	vector<vector<int>> graph(N,init);
	rep(i,M){
		int a, b;
		cin >> a >> b;
		graph[a-1].emplace_back(b-1);
		graph[b-1].emplace_back(a-1);
	}
rep(i,N){
	cout << i<< ": ";
	putv(graph[i]);
}
 
	return graph;
}
void out(bool ans){
	if(ans == true) cout << "yes" << endl;
	else cout << "no" << endl;
}
void Out(bool ans){
	if(ans == true) cout << "Yes" << endl;
	else cout << "No" << endl;
}
void OUT(bool ans){
	if(ans == true) cout << "YES" << endl;
	else cout << "NO" << endl;
}
ll gcd(const ll a, const ll b){
	return (b==0) ? a : gcd(b,a%b);
}
ll lcm(const ll a, const ll b){
	return a*(b/gcd(a,b));
}
ll Manhattan_dist(ll x1, ll y1, ll x2, ll y2){
	return abs(x1-x2) + abs(y1-y2);
}
// nをpで何回割れるか
ll hmt_p_divide_n(ll p, ll n){
	ll s = 0;
	while(n%p == 0){
		n /= p;
		s++;
	}
	return s;
}
int digit(ll n){
	int cntr = 0;
	while(n>0){
		n/=10;
		cntr++;
	}
	return cntr;
}



int main(){
	int N,Q;
	string S;
	cin >>N>>Q>>S;
	// cout << S <<endl;

	// 線形に舐めてあらかじめ答えを全部作っておく．
	// 候補はNC2 = 1/2N(N-1)
	int answer[N][N] = {};
	for(int i=1; i<N; ++i)
		for(int j=0; j<i; ++j)
			answer[j][i] = answer[j][i-1]+(S[i-1]=='A'&&S[i]=='C');
	// rep(i,N){
	// 	rep(j,i)
	// 		cout << answer[j][i] << " ";
	// 	cout << endl;
	// }
	rep(i,Q){
		int l,r;
		cin >> l >> r;
		// cout << "(" << l <<","<<r<<")" <<endl;
		cout << answer[l-1][r-1] << endl;
	}
}