Problem:
Score : 600 points

Problem Statement

In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.

The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.

Given a string S of length N and an integer k (k \geq 3),
he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:

0 \leq a < b < c \leq N - 1

S[a] = D

S[b] = M

S[c] = C

c-a < k

Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.

For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).

Constraints

3 \leq N \leq 10^6

S consists of uppercase English letters

1 \leq Q \leq 75

3 \leq k_i \leq N

All numbers given in input are integers

Input

Input is given from Standard Input in the following format:

N
S
Q
k_{0} k_{1} ... k_{Q-1}

Output

Print Q lines.
The i-th line should contain the k_i-DMC number of the string S.

Sample Input 1

18
DWANGOMEDIACLUSTER
1
18

Sample Output 1

1

(a,b,c) = (0, 6, 11) satisfies the conditions.

Strangely, Dwango Media Cluster does not have so much DMC-ness by his definition.

Sample Input 2

18
DDDDDDMMMMMCCCCCCC
1
18

Sample Output 2

210

The number of triples can be calculated as 6\times 5\times 7.

Sample Input 3

54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40

Sample Output 3

0
1
2

(a, b, c) = (0, 23, 36), (8, 23, 36) satisfy the conditions except the last one, namely, c-a < k_i.

By the way, DWANGO is an acronym for "Dial-up Wide Area Network Gaming Operation".

Sample Output 4

30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20

Sample Output 4

10
52
110
140

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
using namespace std;
 
#define mod 1000000007
#define FOR(x,to) for(int x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define long long long
inline int rei(){int x;cin>>x;return x;}
inline long rel(){long x;cin>>x;return x;}
inline string res(){string x;cin>>x;return x;}
//------------------------------------------------------- 
int ans[100];
int query[100];
int Ms[1000001];
int Ds[1000001];
int Cs[1000001];
long MCs[1000001];
void Calc(){
	int N = rei();
	string S = res();
	int Q = rei();
	for(int i=0;i<Q;i++){
		query[i] = rei(); 
	}
	for(int i=0;i<N;i++){
		if(S[i] == 'D'){
			Ds[i]++;
		}
		if(S[i] == 'M'){
			Ms[i]++;
		}
		if(S[i] == 'C'){
			Cs[i]++;
		}
	}
	for(int i=0;i<N-1;i++){
		Ds[i+1] += Ds[i];
		Ms[i+1] += Ms[i];
		Cs[i+1] += Cs[i];
	}
	for(int i=0;i<N;i++){
		if(S[i] == 'C'){
			MCs[i] = Ms[i];
		}
	}
	for(int i=0;i<N-1;i++){
		MCs[i+1] += MCs[i];
	}
	for(int q=0;q<Q;q++){
		long ans = 0;
		for(int i=0;i<N;i++){
			if(S[i] == 'D'){
				long c = MCs[min(N-1,i+query[q]-1)] - MCs[i];
				c -= ((long)(Cs[min(N-1,i+query[q]-1)] - Cs[i]))*Ms[i];
				ans += c;
			}
		}
		cout << ans << endl;
	}
}
int main(int argc,char** argv){
	ios::sync_with_stdio(false), cin.tie(0);
	cout.tie(0); Calc(); return 0;
}