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)

#include <iostream>
#include<vector>
#include<string>
using namespace std;
typedef long long ll;

int main() {
	int n;cin>>n;
	string s;cin>>s;
	int q;cin>>q;
	vector<int> k(q);
	for(int i=0;i<q;i++){
		cin>>k[i];
	}
	vector<ll> d(n,0);
	vector<ll> m(n,0);
	vector<ll> c(n,0),cm(n,0);
	for(int i=0;i<n;i++){
		if(s[i]=='D'){
			d[i]=1;
		}else if(s[i]=='M'){
			m[i]=1;
		}else if(s[i]=='C'){
			c[i]=1;
		}
	}
//	for(int i=0;i<n;i++){
//		cout<<d[i]<<' ';
//	}
//	cout<<endl;
//	for(int i=0;i<n;i++){
//		cout<<m[i]<<' ';
//	}
//	cout<<endl;
//	for(int i=0;i<n;i++){
//		cout<<c[i]<<' ';
//	}
//	cout<<endl;
	for(int i=1;i<n;i++){
//		d[i]+=d[i-1];
		m[i]+=m[i-1];
		cm[i]+=cm[i-1];
		if(c[i]){
			cm[i]+=m[i];
		}
		c[i]+=c[i-1];
	}
//	for(int i=0;i<n;i++){
//		cout<<cm[i]<<endl;
//	}

	for(int j=0;j<q;j++){
		ll ans=0;
		int tk=k[j];
		for(int i=0;i<n;i++){
			if(s[i]=='D'){
				int r=i+tk-1;
				if(r>=n)r=n-1;
				ll add=cm[r]-cm[i];
				int nc=c[r]-c[i];
				add-=nc*m[i];
				ans+=add;
//				cout<<i<<' '<<add<<' '<<nc<<endl;
			}
		}
		cout<<ans<<endl;
	}
	return 0;
}