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)

fn main() {
    let n: usize = read();
    let p: usize = read();
    let s: String = read();
    let s: Vec<usize> = s.bytes().rev().map(|b| (b - b'0') as usize).collect();

    let mut ans = 0;
    if p == 2 || p == 5 {
        for i in 0..n {
            if s[i] % p == 0 {
                ans += n - i;
            }
        }
        println!("{}", ans);
        return;
    }
    let mut sum = vec![0; n + 1];
    let mut c = 1;
    for i in 0..n {
        sum[i + 1] = sum[i] + c * s[i];
        sum[i + 1] %= p;
        c *= 10;
        c %= p;
    }
    let mut cnt = vec![0; p];
    for i in 0..n + 1 {
        let e = sum[i];
        cnt[e] += 1 as i32;
    }
    for &c in &cnt {
        ans += ((c - 1) * c) as usize / 2;
    }
    println!("{}", ans);
}

use std::io::stdin;
use std::io::Read;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let cin = stdin();
    let cin = cin.lock();
    let token: String = cin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}