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;
#define inf 999999999
#define loop(i, a, b) for (int i = (a), i##len = (b); i < i##len; ++i)
#define rep(i, n) loop(i, 0, n)
#define lin long long
#define lfl long double

lin n;
string s;
lin q;

void solve(lin k)
{
    queue<lin> d;
    queue<lin> m;
    lin ad = 0;
    lin res = 0;

    rep(i, s.length())
    {
        if(d.size() > 0)
        {
            if(i - d.front() >= k)
            {
                d.pop();
                ad -= m.size();

                if(d.size() > 0)
                {
                    while(m.size() > 0 && d.front() > m.front())
                    {
                        m.pop();
                    }
                }
                else
                {
                    while(m.size() > 0)
                    {
                        m.pop();
                    }                    
                }
            } 
        }

        if(s[i] == 'D')
        {
            d.push(i);
        }
        else if(s[i] == 'M')
        {
            if(d.size() > 0)
            {
                m.push(i);
                ad += d.size();
            }
        }
        else if(s[i] == 'C')
        {
            res += ad;
        }
    }

    cout << res << endl;
}

int main(void)
{
    lin k[76];

    cin >> n >> s >> q;
    rep(i, q)
    {
        cin >> k[i];
        solve(k[i]);
    }

    return 0;
}