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)

fn main() {
	let mut s = String::new();
	use std::io::{Read, Write};
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut it = s.split_whitespace();
	let n: usize = it.next().unwrap().parse().unwrap();
	let s: Vec<_> = it.next().unwrap().chars().collect();
	let q: usize = it.next().unwrap().parse().unwrap();
	let out = std::io::stdout();
	let mut out = std::io::BufWriter::new(out.lock());
	for k in it.map(|a| a.parse::<usize>().unwrap()) {
		let mut ans = 0u64;
		let mut cnt = (0, 0, 0);
		for i in 0..s.len() {
			match s[i] {
				'D' => cnt.0 += 1,
				'M' => {
					cnt.1 += 1;
					cnt.2 += cnt.0;
				},
				'C' => {
					ans += cnt.2;
				}
				_ => (),
			}
			if i < k - 1 { continue; }
			match s[i - (k - 1)] {
				'D' => {
					cnt.0 -= 1;
					cnt.2 -= cnt.1;
				}
				'M' => {
					cnt.1 -= 1;
				}
				_ => (),
			}
		}
		writeln!(out, "{}", ans).unwrap();
	}
}