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 Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(non_snake_case,unused_imports)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut buf_it = buf.split_whitespace();

    let N = buf_it.next().unwrap().parse::<usize>().unwrap();
    let S:Vec<_> = buf_it.next().unwrap().chars().collect();
    let Q = buf_it.next().unwrap().parse::<usize>().unwrap();

    let K:Vec<_> = (0..Q).map(|_|
        buf_it.next().unwrap().parse::<usize>().unwrap()
    ).collect();

    let mut D:Vec<_> = (0..N+1).map(|_| 0).collect();
    let mut cntM:Vec<_> = (0..N+1).map(|_| 0).collect();
    let mut sumM:Vec<_> = (0..N+1).map(|_| 0).collect();
    let mut ans:Vec<_> = (0..Q).map(|_| 0).collect();
    for (i,&s) in S.iter().enumerate(){
        D[i+1] = D[i];
        cntM[i+1] = cntM[i];
        sumM[i+1] = sumM[i];
        match s {
            'D' => { D[i+1] += 1;},
            'M' => {
                cntM[i+1] += 1;
                sumM[i+1] += D[i];
            },
            'C' => {
                for (j,&k) in K.iter().enumerate(){
                    let ik = max(0,i as isize - k as isize +1) as usize;
                    let rstD = D[ik];
                    ans[j] += (sumM[i+1]-sumM[ik+1]) - rstD * (cntM[i+1]-cntM[ik+1]);
                }
            },
            _ => {},
        }
    }

/*
    println!("{:?}",D);
    println!("{:?}",cntM);
    println!("{:?}",sumM);
*/

    for &out in &ans{
        println!("{}",out);
    }

}