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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
class Main {
  public static void main (String[] args) throws Exception {
    Scanner sc= new Scanner(System.in);
    
		
     int n = sc.nextInt();
     String s = sc.next();
     int q = sc.nextInt();
     
     int[]k = new int [q];
     
     for(int i=0; i<q; i++){
       k[i]=sc.nextInt();
       }
     
     ArrayList<Integer> d = new ArrayList<>();
     ArrayList<Integer> m = new ArrayList<>();
     ArrayList<Integer> c = new ArrayList<>();
     
     
       
     for(int i=0; i<n; i++){
       if(s.charAt(i)=='D'){
          d.add(i);
       }else if(s.charAt(i)=='M'){
          m.add(i);
       }else if(s.charAt(i)=='C'){
          c.add(i);
       }
      }
      
      int ans =0;
      
    for(int j=0; j<q; j++){
     for(int indexm: m){
       //System.out.println(indexm);
       if(indexm==0)continue;
       if(indexm==n-1)break;
       
       ans = ans+find(d, c, indexm, k[j]);
       }
       //System.out.println(c.get(0));
       System.out.println(ans);
       ans=0;
     }
  }
  
  public static int find(List<Integer> d, List<Integer> c, int indexm, int k){
    
    int ans=0;
    for(int dnum : d){
      if(Math.abs(dnum-indexm)>=k)continue;
      if(dnum>=indexm)break;
         for(int cnum : c){
           if(cnum<= indexm)continue;
           if(Math.abs(cnum-dnum)>=k)break;
 
           ans++;
           //System.out.println("dnum "+ d+" cnum "+c+" ans "+ans);
         }
     }
     return ans;
   }
   
}