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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef int64_t i64;

void run (void) {
  i32 n;
  scanf ("%" SCNi32, &n);
  char *s = (char *) calloc (n + 1, sizeof (char));
  scanf ("%s", s);
  i32 q;
  scanf ("%" SCNi32, &q);
  while (q--) {
    i32 k;
    scanf ("%" SCNi32, &k);
    i64 ans = 0;
    i32 cntD = 0;
    i32 cntM = 0;
    i64 cntDM = 0;
    for (i32 i = 0; i < k; ++i) {
      if (s[i] == 'D') {
	cntD++;
      } else if (s[i] == 'M') {
	cntM++;
	cntDM += cntD;
      } else if (s[i] == 'C') {
	ans += cntDM;
      }
    }
    for (i32 i = k; i < n; ++i) {
      if (s[i - k] == 'D') {
	cntDM -= cntM;
	cntD--;
      } else if (s[i - k] == 'M') {
	cntM--;
      }
      if (s[i] == 'D') {
	cntD++;
      } else if (s[i] == 'M') {
	cntM++;
	cntDM += cntD;
      } else if (s[i] == 'C') {
	ans += cntDM;
      }
    }
    printf("%" PRIi64 "\n", ans);
  }
}

int main (void) {
  run ();
  return 0;
}