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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.to_i
s = gets.chomp!
q = gets.to_i
k = gets.split.map(&:to_i)

cntD = [0]
cntM = [0]
i = 1
s.each_char {|c|
  v1 = c=="D" ? 1 : 0
  v2 = c=="M" ? 1 : 0
  cntD[i] = cntD[i-1] + v1
  cntM[i] = cntM[i-1] + v2
  i += 1
}

cntDM = [0]
1.upto(n) do |i|
  c = s[i-1]
  dm = c == "M" ? cntD[i-1] : 0
  cntDM[i] = cntDM[i-1] + dm
end

k.each do |itv|
  ans = 0
  1.upto(n) do |i|
    if s[i-1] == "C"
      left = (i-itv < 0) ? 1 : i-itv+1
      dmc = cntDM[i-1] - cntDM[left+1] - cntD[left-1] * (cntM[i-1] - cntM[left+1])
      ans += dmc
    end
  end
  p ans
end