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: Time Limit Exceeded)

import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();

        char str[]=in.next().toCharArray();

        char dmc[]=new char[str.length];
        int dmcp[]=new int[str.length];

        int l=0;
        for(int i=0;i<str.length;i++){
            if(str[i]=='D' || str[i]=='M' || str[i]=='C'){
                dmc[l]=str[i];
                dmcp[l]=i;
                l++;
            }
        }

        int q=in.nextInt();
        for(int i=0;i<q;i++){
            int k=in.nextInt();
            int cnt=0;

            for(int d=0;d<l;d++){
                if(dmc[d]=='D'){
                    for(int m=d+1;m<l;m++){
                        if(dmc[m]=='M'){
                            for(int c=m+1;c<l;c++){
                                if(dmc[c]=='C' && dmcp[c]-dmcp[d]<k)cnt++;
                            }
                        }
                    }
                }
            }

            System.out.println(cnt);
        }
    }
}