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: Time Limit Exceeded)

import sys
n = int(input())
s = input()
q = int(input())
k = list(map(int, sys.stdin.readline().split()))
x = []
for i in s:
  if i == "D":
    x.append(1)
  elif i == "M":
    x.append(2)
  elif i == "C":
    x.append(3)
  else:
    x.append(0)
mrui = [0]* (n+1)
for i in range(1,n+1):
  mrui[i] = mrui[i-1]
  if x[i-1] == 2:
    mrui[i] += 1
dp = [[0]*4 for _ in range(n)]
for i in k:
  dp[-1] = [1,0,0,0]
  for j in range(n):
    dp[j] = dp[j-1]
    kon = x[j]
    if kon:
      dp[j][kon] += dp[j-1][kon-1]
    if j >= i-1:
      if x[j-i+1] == 1:
        dp[j][1] -= 1
        dp[j][2] -= (mrui[j+1]-mrui[j-i+2])
  
  print(dp[n-1][3])