Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::Read;

fn run() {
    let mut s = String::new();
    std::io::stdin().read_to_string(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let n: usize = it.next().unwrap().parse().unwrap();
    let p: usize = it.next().unwrap().parse().unwrap();
    let s: Vec<usize> = it.next().unwrap().chars().map(|c| c.to_digit(10).unwrap() as usize).collect();
    if p <= 10 {
        let mut ans = 0;
        let mut dp = vec![0usize; p];
        for &d in s.iter() {
            let mut next = vec![0usize; p];
            for k in 0..p {
                next[(k * 10 + d) % p] += dp[k];
            }
            next[d % p] += 1;
            dp = next;
            ans += dp[0];
        }
        println!("{}", ans);
    } else {
        let mut ans = 0usize;
        let mut now = 0;
        let mut map = std::collections::BTreeMap::new();
        *map.entry(now).or_insert(0) += 1;
        for &d in s.iter() {
            now = (10 * now + d) % p;
            let po = map.entry(now).or_insert(0);
            ans += *po;
            *po += 1;
        }
        println!("{}", ans);
    }
}

fn main() {
    run();
}