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

namespace AtTest.Dwango5
{
    class C
    {
        static void Main(string[] args)
        {
            Method(args);
            Console.ReadLine();
        }

        static void Method(string[] args)
        {
            int n = ReadInt();
            string s = Read();
            int q = ReadInt();
            int[] ks = ReadInts();

            long[] mCnts = new long[n];
            long[] cCnts = new long[n];
            long[] mcCnts = new long[n];
            for (int i = 0; i < n; i++)
            {
                if (i > 0)
                {
                    mCnts[i] = mCnts[i - 1];
                    cCnts[i] = cCnts[i - 1];
                    mcCnts[i] = mcCnts[i - 1];
                }
                if (s[i] == 'M') mCnts[i]++;
                if (s[i] == 'C')
                {
                    cCnts[i]++;
                    mcCnts[i] += mCnts[i];
                }
            }

            for (int i = 0; i < q; i++)
            {
                long res = 0;
                for (int j = 0; j < n; j++)
                {
                    if (s[j] != 'D') continue;

                    int maxI = Math.Min(j + ks[i] - 1, n - 1);
                    res += mcCnts[maxI] - mcCnts[j]
                        - (cCnts[maxI] - cCnts[j]) * mCnts[j];
                }
                Console.WriteLine(res);
            }
        }

        private static string Read() { return Console.ReadLine(); }
        private static int ReadInt() { return int.Parse(Read()); }
        private static long ReadLong() { return long.Parse(Read()); }
        private static double ReadDouble() { return double.Parse(Read()); }
        private static int[] ReadInts() { return Array.ConvertAll(Read().Split(), int.Parse); }
        private static long[] ReadLongs() { return Array.ConvertAll(Read().Split(), long.Parse); }
        private static double[] ReadDoubles() { return Array.ConvertAll(Read().Split(), double.Parse); }
    }
}