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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define int long long

int min(int a, int b){
	return a <= b ? a : b;
}

signed main(){
	int N, Q, K, i, j, q, ans;
	scanf("%lld", &N);
	char *S = (char *)malloc(sizeof(char) * (N + 2));
	scanf("%s", &S[1]);
	int *M_sum = (int *)malloc(sizeof(int) * (N + 1));
	int *C_sum = (int *)malloc(sizeof(int) * (N + 1));
	int *MC_sum = (int *)malloc(sizeof(int) * (N + 1));
	M_sum[0] = 0;
	C_sum[0] = 0;
	MC_sum[0] = 0;
	for(i = 1; i <= N; i++){
		M_sum[i] = M_sum[i - 1] + (S[i] == 'M' ? 1 : 0);
		C_sum[i] = C_sum[i - 1] + (S[i] == 'C' ? 1 : 0);
		MC_sum[i] = MC_sum[i - 1] + (S[i] == 'C' ? M_sum[i] : 0);
	}
/*	printf("M_sum:");
	for(i = 0; i <= N; i++){
		printf("%lld ", M_sum[i]);
	}
	printf("\n");
	printf("C_sum:");
	for(i = 0; i <= N; i++){
		printf("%lld ", C_sum[i]);
	}
	printf("\n");
	printf("MC_sum:\n");
	for(i = 0; i <= N; i++){
		printf("%lld ", MC_sum[i]);
	}
	printf("\n");
*/	scanf("%lld", &Q);
	for(q = 0; q < Q; q++){
		scanf("%lld", &K);
		for(i = 1, ans = 0; i <= N; i++){
			if(S[i] == 'D'){
				j = min(N, i + K - 1);
				ans += (MC_sum[j] - MC_sum[i - 1]) - (C_sum[j] - C_sum[i - 1]) * M_sum[i - 1];
			}
		}
		printf("%lld\n", ans);
	}
	return 0;
}