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 (3.4.3) (judge verdict: Time Limit Exceeded)

#! length of string
length_input = input()
length = int(length_input)
#! content of string
test_str = input()
#! number of gaps
gaps_num_input = input()
gaps_num = int(gaps_num_input)
#! gaps
gaps_str = input()
#! get the gaps list
gaps_list = gaps_str.split(" ")
count = 0

list_D = []
list_M = []
list_C = []

#! get all the indexs of D,M,C
for char in test_str:
    if char == "D" :
        list_D.append(count)
        count += 1        
    elif char == "M" :    
        list_M.append(count)
        count += 1
    elif char == "C" :
        list_C.append(count)
        count += 1
    else :
        count += 1
        continue    
   
for gap_str in gaps_list:
    n = 0
    gap = int(gap_str)
    for i in list_D :
        for j in list_M :
            if i < j :
                for k in list_C :
                    if j < k :
                        if k-i < gap :
                            n += 1
                        else :
                            break
                    else :
                        continue
            else :
                continue
    
    print(n)