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 <bits/stdc++.h>
using namespace std;

using int32 = int_fast32_t;
using int64 = int_fast64_t;

const int32 INF = 1e9;
const int32 MOD = 1e9+7;
const int64 LLINF = 1e18;

#define YES(n) cout << ((n) ? "YES\n" : "NO\n"  )
#define Yes(n) cout << ((n) ? "Yes\n" : "No\n"  )
#define ANS(n) cout << (n) << "\n"
#define REP(i,n) for(int64 i=0;i<(n);++i)
#define FOR(i,a,b) for(int64 i=(a);i<(b);i++)
#define FORR(i,a,b) for(int64 i=(a);i>=(b);i--)
#define ALL(obj) (obj).begin(),(obj).end()
#define pii pair<int32,int32>
#define pll pair<int64,int64>
#define pb(a) push_back(a)
#define mp make_pair

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int32 n;
    cin >> n;
    string s;
    cin >> s;
    int32 q;
    cin >> q;
    while(q--){
        int32 k;
        cin >> k;
        int64 res = 0;
        int32 d = 0, m = 0, c = 0;
        int64 comb = 0;
        REP(i,n){
            if(i - k >= 0){
                if(s[i-k] == 'D'){
                    d--;
                    comb -= m;
                }else if(s[i-k] == 'M'){
                    m--;
                }else if(s[i-k] == 'C'){
                    c--;
                }
            }
            if(s[i] == 'D'){
                d++;
            }else if(s[i] == 'M'){
                m++;
                comb += d;
            }else if(s[i] == 'C'){
                c++;
                res += comb;
            }
            // cout << d << " " << m << " " << c << " " << comb << endl;
        }
        cout << res << endl;
    }
    return 0;
}