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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;

public class Main {

    private int n;
    private String s;
    private int q;
    private int[] k;
    private int len;
    private List<Integer> a;
    private List<Integer> b;
    private List<Integer> c;
    
    public static void main(String[] args) {
	Main m = new Main();
	m.conduct();
    }

    private void conduct() {
	try (BufferedReader br = new BufferedReader(new InputStreamReader(System.in))) {
	    n = Integer.parseInt(br.readLine());
	    s = br.readLine();
	    q = Integer.parseInt(br.readLine());
	    k = Arrays.stream(br.readLine().split("[\\s]+")).mapToInt(Integer::parseInt).toArray();
	    a = new ArrayList<>();
	    b = new ArrayList<>();
	    c = new ArrayList<>();
	} catch(IOException e) {
	    e.printStackTrace();
	}
	
	for (int i=0;i<n;i++) {
	    switch (s.charAt(i)) {
	    case 'D':
		a.add(i);
		break;
	    case 'M':
		b.add(i);
		break;
	    case 'C':
		c.add(i);
		break;
	    }
	}

	int[] m = new int[q];
	
	for (int ia : a) {
	    Iterator<Integer> itb = b.iterator();
	    while (itb.hasNext()) {
		int ib = itb.next();
		if (ib < ia) {
		    itb.remove();
		    continue;
		}
		Iterator<Integer> itc = c.iterator();
		while (itc.hasNext()) {
		    int ic = itc.next();
		    if (ic < ia) {
			itc.remove();
			continue;
		    } else if (ic < ib) {
			continue;
		    }
		    for (int i=0;i<q;i++) {
			if (ic-ia < k[i]) m[i]++;
		    }
		}
	    }
	}
	
	for (int i=0;i<q;i++) System.out.println(m[i]);
	
    }

}