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 (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused, non_snake_case, dead_code, non_upper_case_globals)]
use {
    proconio::{marker::*, *},
    std::{cmp::*, collections::*, mem::*},
};
#[fastout]
fn main() {
    input! {
        n:usize,p:usize,
        s:Chars,
    }
    let mut s = s;
    if p == 2 || p == 5 {
        let mut ans = 0i64;
        for (i, &c) in s.iter().enumerate() {
            if char_conv(c, '0') % p == 0 {
                ans += (i + 1) as i64;
            }
        }
        println!("{}", ans);
        return;
    }
    s.reverse();
    let mut tens = 1usize;
    let mut nm = 0;
    let mut ls = vec![0; p + 3];
    ls[0] = 1;
    for (i, &c) in s.iter().enumerate() {
        let now = char_conv(c, '0');
        nm = (now * tens + nm) % p;
        tens = (tens * 10) % p;
        ls[nm] += 1;
    }
    let ans = ls.iter().fold(0, |sum, &cnt| sum + cnt * (cnt - 1) / 2);
    println!("{}", ans);
}
pub fn char_conv(c: char, base: char) -> usize {
    return (c as u8 - base as u8) as usize;
}