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: Wrong Answer)

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

namespace test
{
    class Program
    {
        static void Main(string[] args)
        {
            var N = ReadInt();
            var S = Read();
            var Q = ReadInt();
            var k = ReadArrayInt();
            var D = new int[N + 1];
            var M = new int[N + 1];
            var DM = new long[N + 1];
            for (int i = 0; i < N; i++)
            {
                if (S[i] == 'D')
                {
                    D[i + 1] = D[i] + 1;
                    M[i + 1] = M[i];
                    DM[i + 1] = DM[i];
                }
                else if (S[i] == 'M')
                {
                    D[i + 1] = D[i];
                    M[i + 1] = M[i] + 1;
                    DM[i + 1] = DM[i] + D[i];
                }
                else
                {
                    D[i + 1] = D[i];
                    M[i + 1] = M[i];
                    DM[i + 1] = DM[i];
                }
            }
            foreach(var kk in k)
            {
                var ans = 0L;
                for (int i = 0; i < N; i++)
                {
                    if (S[i] == 'C')
                    {
                        ans += DM[i];
                        if (i >= kk)
                        {
                            ans -= D[i - kk + 1] * (M[i] - M[i - kk + 1]);
                            ans -= DM[i - kk + 1];
                        }
                    }
                }
                Console.WriteLine(ans);
            }

            Console.WriteLine();
            Console.ReadLine();
        }

        static string Read() { return Console.ReadLine(); }
        static int ReadInt() { return int.Parse(Read()); }
        static long ReadLong() { return long.Parse(Read()); }
        static int[] ReadArrayInt() { return Read().Split(' ').Select(s => int.Parse(s)).ToArray(); }
        static long[] ReadArrayLong() { return Read().Split(' ').Select(s => long.Parse(s)).ToArray(); }
    }
}