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: Wrong Answer)

#include <iostream>
#include <sstream>
#include <algorithm>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <memory>
#include <complex>
#include <numeric>
#include <cstdio>
#include <iomanip>

#define REP(i,m,n) for(int i=int(m);i<int(n);i++)
#define RREP(i,m,n) for(int i=int(n)-1;i>=int(m);--i)
#define EACH(i,c) for (auto &(i): c)
#define all(c) begin(c),end(c)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort(begin(c),end(c))
#define pb emplace_back
#define MP make_pair
#define SZ(a) int((a).size())

#ifdef LOCAL
#define DEBUG(s) cout << (s) << endl
#define dump(x)  cerr << #x << " = " << (x) << endl
#define BR cout << endl;
#else
#define DEBUG(s) do{}while(0)
#define dump(x) do{}while(0)
#define BR 
#endif
using namespace std;

using UI = unsigned int;
using UL = unsigned long;
using LL = long long int;
using ULL = unsigned long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VLL = vector<LL>;
using VS = vector<string>;
using PII = pair<int,int>;
using VP = vector<PII>;

//struct edge {int from, to, cost;};

constexpr double EPS = 1e-10;
//constexpr double PI  = acos(-1.0);
//constexpr int INF = INT_MAX;
constexpr int MOD = 1'000'000'007;

template<class T> inline T sqr(T x) {return x*x;}

void solve() {
    int n,q;
    string s;
    cin >> n >> s >> q;
    int len = s.size();
    VLL k(q);
    REP(i,0,q) cin >> k[i];
    /*
    VVI cnt(3, VI(len));
    VVI ru(3, VI(len+1));
    REP(i,0,len) {
        if (s[i] == 'D') ++cnt[0][i];
        if (s[i] == 'M') ++cnt[1][i];
        if (s[i] == 'C') ++cnt[2][i];
    }
    REP(i,0,len) {
        REP(j,0,3) ru[j][i+1] = ru[j][i] + cnt[j][i];
    }
    */
    VI ar(len+1);
    {
        VI cnt(len);
        REP(i,0,len) if (s[i] == 'D') ++cnt[i];
        REP(i,0,len) ar[i+1] = ar[i] + cnt[i];
    }
   VI br, bp, cp;
   {
       int cnt = 0;
       REP(i,0,len) {
            if (s[i] == 'D') ++cnt;
            if (s[i] == 'M') {
               br.push_back(cnt);
               bp.push_back(i);
            }
            if (s[i] == 'C') {
                cp.push_back(i);
            }
       }
    }
    VI brr(br.size()+1);
    REP(i,0,br.size()) brr[i+1] = brr[i] + br[i];
    //REP(i,0,brr.size()) dump(brr[i]);
    REP(i,0,q) {
        LL ans = 0;
        for (auto &c: cp) {
            auto rt = upper_bound(all(bp), c);
            auto lt = lower_bound(all(bp), c - k[i] + 1);
            int d = rt - lt;
            ans += brr[rt-bp.begin()] - brr[lt-bp.begin()] - d * ar[max(0LL,c - k[i] + 1)];
            //if (k[i] != 40) continue;
            //dump(c);
            //dump(*(rt-1));
            //dump(*lt);
            //dump(d);
            //dump(brr[rt-bp.begin()]);
            //dump(brr[lt-bp.begin()]);
            //dump(d * ar[max(0LL,c - k[i] + 1)]);
        }
        cout << ans << endl;
    } 
}

int main() {
    solve();
    
    return 0;
}