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++14 (GCC 5.4.1) (judge verdict: Accepted)

//cf.cpp
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <vector>
#include <string.h>
#include <cstring>
#include <queue>
#include <math.h>
#include <stack>
#include <bitset>
#include <utility>
#define mp make_pair
#define pb push_back
#define llc (long long)
#define pob pop_back
#define pm %1000000007
using namespace std;

bool comp(long long x, long long y){
	return x>y;
}
int main(){
	int cuf;
	int m=0, k;
	vector <int> d, c;
	vector <int> dl, cl;
	queue <int> csa;
	long long tot=0;
	int n, q;
	cin>>n;
	getchar();
	long long ans;
	for(int i = 0;i < n;i ++){
		switch(getchar()){
			case 'D':
			d.pb(i);
			dl.pb(m);
			break;
			case 'M':
			m++;
			break;
			case 'C':
			c.pb(i);
			cl.pb(m);
			break;
		}
	}
	cin>>q;
	while(q--){
		while(!csa.empty())csa.pop();
		cin>>k;
		ans=0;
		tot=0;
		for(int i = 0;i < c.size();i ++){
			if(d[0]+k<=c[i]){
				cuf=i;
				break;
			}
			if(d[0]<c[i]){
				csa.push(i);
				tot+=cl[i];
			}
			if(i+1==c.size()){
				cuf=c.size();
			}
		}
		ans+=tot-llc dl[0]*csa.size();
		for(int i = 1;i < d.size();i ++){
			for(int j = cuf;j < c.size();j ++){
				if(d[i]+k<=c[j]){
					cuf=j;
					break;
				}
				if(d[i]<c[j]){
					csa.push(j);
					tot+=cl[j];
				}
				if(j+1==c.size()){
					cuf=c.size();
				}
			}
			while(!csa.empty()){
				if(d[i]>c[csa.front()]){
					tot-=cl[csa.front()];
					csa.pop();
				}else{
					break;
				}
			}
			ans+=tot-llc dl[i]*csa.size();
		}
		printf("%lld\n",ans);
	}
}