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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define debug(var) do{cout << #var << " : "; view(var);}while(0)
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
template<class T> void view(T e) {cout << e << endl;}
template<class T> void view(const vector<T> &v) {for(const auto &e : v){cout << e << " ";} cout << endl;}
template<class T> void view(const vector<vector<T>> &vv) {for(const auto &v : vv){view(v);}}
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9+10;
const ll inf_l = 1e18;
const int MAX = 1e5;

int main() {
    int n; cin >> n;
    string s; cin >> s;
    int q; cin >> q;
    vint k(q);
    rep(i,q) cin >> k[i];
    int ct_d = 0;
    vll dp_dm(n+1,0);
    vint sum_d(n+1,0), sum_m(n+1,0);
    REP(i,n+1) {
        dp_dm[i] += dp_dm[i-1];
        sum_d[i] += sum_d[i-1];
        sum_m[i] += sum_m[i-1];
        if (s[i-1] == 'D') {
            ct_d++;
            sum_d[i]++;
        }
        if (s[i-1] == 'M') {
            dp_dm[i] += ct_d;
            sum_m[i]++;
        }
    }
    rep(i,q) {
        ll ans = 0;
        int r = 0;
        while (r < k[i]) {
            if (s[r] == 'C') ans += dp_dm[r];
            r++;
        }
        ll sum_dm = dp_dm[r];
        while (r < n) {
            if (s[r-k[i]] == 'D') sum_dm -= sum_m[r] - sum_m[r-k[i]];
            if (s[r] == 'M') sum_dm += sum_d[r] - sum_d[r-k[i]];
            if (s[r] == 'C') ans += sum_dm;
            r++; 
        }
        cout << ans << endl;
    }
}