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 Python (3.8.2) (judge verdict: Wrong Answer)

import numpy as np
from numba import njit

N = int(input())
S = input()
Q = int(input())
ks = list(map(int, input().split()))

def func(ch):
    if ch == 'D':
        return 1
    if ch == 'M':
        return 2
    if ch == 'C':
        return 3
    return 0
    
S_arr = np.array([func(ch) for ch in S], dtype = np.int32)

@njit("i4(i4, i4[:], i4)", cache = True)
def solve(N, S_arr, k):
    ans = 0
    D = 0
    M = 0
    DM = 0
    for i in range(k):
        if S_arr[i] == 1:
            D += 1
        elif S_arr[i] == 2:
            DM += D
            M += 1
        elif S_arr[i] == 3:
            ans += DM
    for i in range(k, N):
        if S_arr[i-k] == 1:
            D -= 1
            DM -= M
        elif S_arr[i-k] == 2:
            M -= 1

        if S_arr[i] == 1:
            D += 1
        elif S_arr[i] == 2:
            DM += D
            M += 1
        elif S_arr[i] == 3:
            ans += DM

    return ans

for k in ks:
    print(solve(N, S_arr, k))