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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.Text;

class TEST{
	static void Main(){
		Sol mySol =new Sol();
		mySol.Solve();
	}
}

class Sol{
	public void Solve(){
		
		foreach(var k in K){
			long ans = 0;
			long ms = 0;
			long cs = 0;
			long mcs = 0;
			int l = 0;
			int r = 0;
			for(r = 0;r < k;r++){
				if(S[r] == 'M'){
					ms++;
				}
				if(S[r] == 'C'){
					mcs += ms;
					cs++;
				}
			}
//Console.WriteLine("ms:{0} mcs:{1} cs:{2}",ms, mcs, cs);			
			for(l=0;l<N;l++){
				if(S[l] == 'D'){
					ans += mcs;
				}
				
				if(S[l] == 'M'){
					mcs -= cs;
					ms--;
				}
				if(S[l] == 'C'){
					cs--;
				}
				
				if(r < N){
					if(S[r] == 'M'){
						ms++;
					}
					if(S[r] == 'C'){
						mcs += ms;
						cs++;
					}
					r++;
				}
//Console.WriteLine("ms:{0} mcs:{1} cs:{2}",ms, mcs, cs);			
			}
			
			Console.WriteLine(ans);
		}
		
	}
	int N;
	String S;
	int Q;
	int[] K;
	public Sol(){
		N = ri();
		S = rs();
		Q = ri();
		K = ria();
	}

	static String rs(){return Console.ReadLine();}
	static int ri(){return int.Parse(Console.ReadLine());}
	static long rl(){return long.Parse(Console.ReadLine());}
	static double rd(){return double.Parse(Console.ReadLine());}
	static String[] rsa(char sep=' '){return Console.ReadLine().Split(sep);}
	static int[] ria(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>int.Parse(e));}
	static long[] rla(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>long.Parse(e));}
	static double[] rda(char sep=' '){return Array.ConvertAll(Console.ReadLine().Split(sep),e=>double.Parse(e));}
}



class BIT{
	int MM;
	int n;
	long[] bit;
	public BIT(int n_){
		n = n_;
		MM = 1 << 2;
		while(MM < n * 2) MM <<= 1;
		bit = new long[MM + 1];
	}
	//1-origin!!
	public void Add(int i, long x){
		while(i <= n){
			bit[i] += x;
			i += (i&-i);
		}
	}
	
	// i=>sum(a_1,...,a_i)
	// i=0 -> return 0;
	public long Sum(int i){
		long s = 0;
		while(i > 0){
			s += bit[i];
			i -= (i&-i);
		}
		return s;
	}
}