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::*;
use std::str::FromStr;
use utils::*;

pub fn main() {
    let i = stdin();
    let mut o = Vec::new();
    run(i.lock(), &mut o);
    stdout().write_all(&o).unwrap();
}

fn run<R: BufRead, W: Write>(i: R, o: &mut W) {
    let mut i = CpReader::new(i);
    let _n = i.read::<usize>();
    let s = i.read_line().to_string();
    let q = i.read::<usize>();
    let k = i.read_vec::<usize>(q);
    for k in k {
        writeln!(o, "{}", solve(&s, k)).unwrap();
    }
}
pub fn solve(s: &str, k: usize) -> usize {
    let s = s.as_bytes();
    let mut d = 0;
    let mut m = 0;
    let mut dm = 0;

    let mut result = 0;
    let mut j = 0;

    for i in 0..s.len() {
        if i - j == k {
            match s[j] {
                b'D' => {
                    d -= 1;
                    dm -= m;
                }
                b'M' => {
                    m -= 1;
                }
                _ => {}
            }
            j += 1;
        }
        match s[i] {
            b'D' => {
                d += 1;
            }
            b'M' => {
                m += 1;
                dm += d;
            }
            b'C' => {
                result += dm;
            }
            _ => {}
        }
    }
    result
}

mod utils {
    use super::*;

    pub struct CpReader<R: BufRead> {
        r: R,
        s: String,
    }

    macro_rules! fn_read {
        { $f:ident($($t:ident),*) } => {
            #[allow(dead_code)]
            pub fn $f<$($t: FromStr),*>(&mut self) -> ($($t),*) {
                let i = &mut self.read_line().split(' ');
                ($(read_next::<$t,_>(i)),*)
            }
        };
    }

    impl<R: BufRead> CpReader<R> {
        pub fn new(r: R) -> Self {
            CpReader {
                r: r,
                s: String::new(),
            }
        }
        pub fn read_line(&mut self) -> &str {
            self.s.clear();
            self.r.read_line(&mut self.s).unwrap();
            self.s.trim()
        }

        pub fn read<T: FromStr>(&mut self) -> T {
            self.read_line().parse().ok().unwrap()
        }
        fn_read! { read2(T1, T2) }
        fn_read! { read3(T1, T2, T3) }
        fn_read! { read4(T1, T2, T3, T4) }
        fn_read! { read5(T1, T2, T3, T4, T5) }

        pub fn read_vec<T: FromStr>(&mut self, n: usize) -> Vec<T> {
            self.read_line()
                .split(' ')
                .take(n)
                .map(|x| x.parse().ok().unwrap())
                .collect()
        }
    }
    fn read_next<'a, T: FromStr, I: Iterator<Item = &'a str>>(i: &mut I) -> T {
        i.next().unwrap().parse().ok().unwrap()
    }

}