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: Time Limit Exceeded)

#![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 A = HashSet::new();
    for (i,&s) in S.iter().enumerate(){
        if s == 'D'{
            A.insert(i as isize);
        }
    }

    for &k in &K{
        let k = k as isize;
        let mut ans = 0;
        let mut B = HashMap::new();
        let mut cntA = 0;
        let mut rstA = 0;
        let mut cntB = 0;
        let mut sumB = 0;
        for i in 0..N{
            let i = i as isize;
            if A.contains(&i){
                cntA += 1;
            }
            if A.contains(&(i-k)){
                rstA += 1;
            }
            if B.contains_key(&(i-k)){
                cntB -= 1;
                sumB -= *B.get(&(i-k)).unwrap();
            }
            if S[i as usize] == 'M'{
                cntB += 1;
                sumB += cntA;
                B.insert(i,cntA);
            }
            
            if S[i as usize] == 'C'{
                ans += sumB - rstA * cntB;
            }
        }
        println!("{}",ans);
    }
    

}