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);
        }
    }

    let mut D = vec![0];
    let mut cntM = SegmentTree::new(N+1,f,0);
    let mut sumM = SegmentTree::new(N+1,f,0);
    let mut ans:Vec<_> = (0..Q).map(|_| 0).collect();
    for (i,&s) in S.iter().enumerate(){
        let d = D[i];
        D.push(d + if s=='D' {1} else {0});
        if s=='M'{
            cntM.set(i+1,1);
            sumM.set(i+1,D[i]);
        }
        if s=='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.get(ik+1,i+1) - rstD * cntM.get(ik+1,i+1);
            }
        }
    }

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

}

fn f(x:isize,y:isize)->isize{
    x+y
}

struct SegmentTree<N>{
    n: usize,
    T: Vec<N>,
    func: fn(N,N) -> N,
    identity: N,
}

impl<N:Copy> SegmentTree<N> {
    fn new(n:usize,func:fn(N,N) -> N,identity:N) -> SegmentTree<N>{
        let mut T = SegmentTree { n:n,T:Vec::new(),func:func,identity:identity};
        T.init();
        T
    }
    fn init(&mut self){
        self.n = 1<<format!("{:b}",self.n-1).len();
        self.T = (0..(2*self.n-1)).map(|_| self.identity).collect();
    }
    fn set(&mut self,index:usize,value:N){
        let i = index + self.n - 1;
        self.T[i] = value;
        self._update_parent((i-1)/2);
    }
    fn _update_parent(&mut self,i:usize){
        self.T[i] = (self.func)(self.T[i*2+1],self.T[i*2+2]);
        if i>0 { self._update_parent((i-1)/2)}
    }
    fn get(&self,l:usize,r:usize) -> N{
    self._get(l,r,0,0,self.n)
    }
    fn _get(&self,l:usize,r:usize,i:usize,a:usize,b:usize) -> N{
        if l<=a && b<=r {
            self.T[i]
        } else if b<=l || r<=a {
            self.identity
        } else{
            (self.func)(self._get(l,r,2*i+1,a,(a+b)/2),
                        self._get(l,r,2*i+2,(a+b)/2,b))
        }
    }
}