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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,q;
    string s;
    cin >> n >> s >> q;
    vector<int> k(q),d(0),m(0),c(0),sub(0);
    for(int i = 0;i < q;i ++){
        cin >> k[i];
    }
    for(int i = 0;i < n;i ++){
        if(s[i]=='D') d.push_back(i);
        if(s[i]=='M') m.push_back(i);
        if(s[i]=='C') c.push_back(i);
    }
    for(int i = 0;i < d.size();i ++){
        for(int j = 0;j < c.size();j ++){
            for(int l = 0;l < m.size();l ++){
                if(m[l]>d[i]&&m[l]<c[j]) sub.push_back(c[j]-d[i]);
            }
        }
    }
    for(int i = 0;i < q;i ++){
        int ans = 0;
        for(int j = 0;j < sub.size();j ++){
            if(sub[j]<k[i]) ans ++;
        }
        cout << ans << endl;
    }
    return 0;
}