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: Time Limit Exceeded)

#include<stdio.h>

char c;
int f;
int nextInt();

int N, Q, i, j, k;
long answer;
char S[1000000];
int K;
int M[1000001];

int main(void){
	N = nextInt();
	M[0] = 0;
	for(i=0; i<N; i++){
		S[i] = getchar();
		M[i+1] = (S[i]=='M' ? M[i]+1: M[i]);
	}
	Q = nextInt();
	for(i=0; i<Q; i++){
		K = nextInt();
		answer = 0;
		for(j=0; j<N; j++){
			if(S[j] == 'D'){
				for(k=j+2; k<N&&k<K+j; k++){
					if(S[k] == 'C'){
						answer += M[k] - M[j+1];
					}
				}
			}
		}
		printf("%ld\n", answer);
	}
	return 0;
}

int nextInt(){
	f=0;
	c = getchar();
	while((c<'0')||('9'<c)){
		c=getchar();
	}
	while((c>='0')&&(c<='9')){
		f=f*10+(c-'0');
		c=getchar();
	}
	return f;
}