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: Accepted)

use std::io::Read;
use std::io::Write;

fn run() {
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    it.next();
    let s: Vec<char> = it.next().unwrap().chars().collect();
    it.next();
    for k in it {
        let k: usize = k.parse().unwrap();
        let mut d = 0u64;
        let mut m = 0;
        let mut dm = 0;
        let mut dmc = 0;
        for &c in s[..k].iter() {
            if c == 'D' {
                d += 1;
            } else if c == 'M' {
                m += 1;
                dm += d;
            } else if c == 'C' {
                dmc += dm;
            }
        }
        for (&f, &t) in s.iter().zip(s[k..].iter()) {
            if f == 'D' {
                dm -= m;
                d -= 1;
            } else if f == 'M' {
                m -= 1;
            }
            if t == 'D' {
                d += 1;
            } else if t == 'M' {
                m += 1;
                dm += d;
            } else if t == 'C' {
                dmc += dm;
            }
        }
        writeln!(out, "{}", dmc).ok();
    }
}

fn main() {
    run();
}