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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;

typedef long long ll;
const int maxn = 1e6 + 7;
#define sz(x) (int)(x).size()

int k[100];

struct FT {
	vector<ll> s;
	FT(int n) : s(n) {}
	void update(int pos, ll dif) { // a[pos] += dif
		for (; pos < sz(s); pos |= pos + 1) s[pos] += dif;
	}
	ll query(int pos) { // sum of values in [0, pos)
		ll res = 0;
		for (; pos > 0; pos &= pos - 1) res += s[pos-1];
		return res;
	}
	int lower_bound(ll sum) {// min pos st sum of [0, pos] >= sum
		// Returns n if no sum is >= sum, or -1 if empty sum is.
		if (sum <= 0) return -1;
		int pos = 0;
		for (int pw = 1 << 25; pw; pw >>= 1) {
			if (pos + pw <= sz(s) && s[pos + pw-1] < sum)
				pos += pw, sum -= s[pos-1];
		}
		return pos;
	}
};

string str;
int n, q, numOfm = 0;

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	cin >> str;
	cin >> q;
	for (int i = 0; i < q; i++)
		cin >> k[i];
	FT cnt(n+10), alpha(n+10);
	for (int i = 0; i < n; i++) {
		if (str[i] == 'C') {
			alpha.update(i, 1);
			cnt.update(i, numOfm);
		} else if (str[i] == 'M') {
			numOfm++;
		}
	}
	for (int idx = 0; idx < q; idx++) {
		ll res = 0;
		numOfm = 0;
		for (int i = 0; i < n; i++) {
			if (str[i] == 'M') numOfm++;
			if (str[i] == 'D') {
				ll numOfc = alpha.query(i+k[idx])-alpha.query(i);
				ll sumOfc = cnt.query(i+k[idx])-cnt.query(i);
				res += (sumOfc - (numOfm * numOfc));
			}
		}
		cout << res << endl;
	}
}