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: Time Limit Exceeded)

N = gets.to_i
S = gets.chomp.split('')
Q = gets.to_i
K = gets.split.map(&:to_i)

require 'pp'
def debug obj
  # PP.pp obj, STDERR
end

STRD = 'D'.freeze
STRM = 'M'.freeze
STRC = 'C'.freeze
d_list = []
m_sum = Array.new(N, 0)
c_sum = Array.new(N, 0)

S.each_with_index do |s, i|
  d_list.push(i) if s == STRD
  m_sum[i] = m_sum[i-1] + (s == STRM ? 1 : 0)
  c_sum[i] = c_sum[i-1] + (s == STRC ? 1 : 0)
end
# debug m_sum
# debug c_sum
Q.times do |q|
  k = K[q]
  result = 0
  d_list.each do |d|
    from = d
    to = d+k-1
    to = ( N - 1) if to > ( N - 1)
    while(c = (from..to).bsearch{|i| c_sum[i] > c_sum[from] }) do
      result += m_sum[c] - m_sum[d]
      debug [d, c, result]
      from = c
    end
  end
  puts result
end