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 AtCoderCS
{
	public interface IReader
	{
		string ReadLine();
		int ReadInt();
		long ReadLong();
		string[] ReadStringArray();
		int[] ReadIntArray();
		long[] ReadLongArray();
	}
	public class ConsoleReader : IReader
	{
		public string ReadLine() => Console.ReadLine();
		public int ReadInt() => int.Parse(ReadLine());
		public long ReadLong() => long.Parse(ReadLine());
		public string[] ReadStringArray() => ReadLine().Split(' ');
		public int[] ReadIntArray() => ReadLine().Split(' ').Select<string, int>(s => int.Parse(s)).ToArray();
		public long[] ReadLongArray() => ReadLine().Split(' ').Select<string, long>(s => long.Parse(s)).ToArray();
	}

	public interface IWriter
	{
		void WriteLine(string line);
		void WriteLine(int value);
		void WriteLine(ulong value);
	}
	public class ConsoleWriter : IWriter
	{
		public void WriteLine(string line) => Console.WriteLine(line);
		public void WriteLine(ulong value) => Console.WriteLine(value);
		public void WriteLine(int value) => Console.WriteLine(value);
	}

	public partial class Solver
	{
		readonly IReader _reader;
		readonly IWriter _writer;

		public Solver(IReader reader = null, IWriter writer = null) {
			_reader = reader ?? new ConsoleReader();
			_writer = writer ?? new ConsoleWriter();
		}

		partial void Dump<T>(IEnumerable<T> array);

		public void Run() {
			var N = _reader.ReadInt();
			//Console.WriteLine(N);

			var S = _reader.ReadLine();

			// D/M数, DM組数を累積和で保持する ★i番目の前までの和 を格納する
			ulong[] accumD = new ulong[S.Length];
			ulong[] accumM = new ulong[S.Length];
			ulong[] accumDM = new ulong[S.Length];
			ulong sumD = 0, sumM = 0, sumDM = 0;
			for (int i = 0; i < S.Length; i++) {
				accumD[i] = sumD;
				accumM[i] = sumM;
				accumDM[i] = sumDM;
				switch (S[i]) {
					case 'D': ++sumD; break;
					case 'M':
						sumDM += sumD;
						++sumM;
						break;
				}
			}
			_reader.ReadInt();

			int[] k_ary = _reader.ReadIntArray();
			//k_ary.Dump();

			foreach (var k in k_ary) {
				int sect_len = k - 1;
				int l = 0;
				ulong ans = 0;

				// 右端のindexをインクリメントしていく
				for (int r = 2; r < N; r++) {
					// 左端
					if (S[r] == 'C') {
						if (0 <= r - sect_len) l = r - sect_len;
						ans += accumDM[r] - accumDM[l] - accumD[l] * (accumM[r] - accumM[l]);
					}
				}
				_writer.WriteLine(ans);
			}
		}
	}

	class MainClass
	{
		public static void Main(string[] args) {
			new Solver().Run();
		}
	}
}