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 PyPy3 (2.4.0) (judge verdict: Accepted)

n = int(input())
s = input()
acc_cnt_a = [0] * n
acc_cnt_b = [0] * n
acc_cnt_ab = [0] * n
c_loc = []
cnt_a = 0
cnt_b = 0
cnt_ab = 0
for i in range(n):
    c = s[i]
    if c == 'D':
        cnt_a += 1
    elif c == 'M':
        cnt_b += 1
        cnt_ab += cnt_a
    elif c == 'C':
        c_loc.append(i)
    acc_cnt_a[i] = cnt_a
    acc_cnt_b[i] = cnt_b
    acc_cnt_ab[i] = cnt_ab

q = int(input())
buf = []
for k in map(int, input().split()):
    ans = 0
    for r in c_loc:
        l = r - k
        if l >= 0:
            rejected_a = acc_cnt_a[l]
            cnt_internal_b = acc_cnt_b[r] - acc_cnt_b[l]
            cnt_internal_ab = acc_cnt_ab[r] - acc_cnt_ab[l]
            ans += cnt_internal_ab - rejected_a * cnt_internal_b
        else:
            ans += acc_cnt_ab[r]
    buf.append(ans)
print('\n'.join(map(str, buf)))