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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;
public class Main {
	public static int[] qlist;

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		char[] s = sc.next().toCharArray();
		int q = sc.nextInt();
		qlist = new int[q];
		int dcnt = 0,dmcnt = 0,cnt = 0,mcnt = 0,k;
		char c;
		for(int i = 0;i < q;i++) {
			qlist[i] = sc.nextInt();
		}
		for(int j = 0;j < q;j++) {
			dmcnt = 0;
			dcnt = 0;
			mcnt = 0;
			cnt = 0;
			for(int i = 0;i < n;i++) {
				k = qlist[j];
				if(i - k >= 0) {
					c = s[i-k];
					if(c == 'D') {
						dcnt--;
						dmcnt -= mcnt;
					}
					else if(c == 'M') {
						mcnt--;
					}
				}
				if(s[i] == 'D') {
					dcnt++;
				}
				else if(s[i] == 'M') {
					mcnt++;
					dmcnt += dcnt;
				}
				else if(s[i] == 'C') {
					cnt += dmcnt;
				}
			}
			System.out.println(cnt);
		}

	}
}