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 main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut it = buf.split_whitespace();
    let _n = it.next().unwrap().parse::<i64>().unwrap();
    let p = it.next().unwrap().parse::<i64>().unwrap();
    let s = it.next().unwrap();
    let s_it = s.bytes().map(|c| (c - b'0') as i64);

    if p == 2 || p == 5 {
        let mut cnt = 0;
        for (i, digit) in s_it.enumerate() {
            if digit % p == 0 {
                cnt += i+1;
            }
        }
        println!("{}", cnt);
        return;
    }


    let mut mods = vec![0; p as usize];
    mods[0] = 1;

    let mut prev = 0;
    let mut exp = 1;
    for digit in s_it.rev() {
        let current = (prev + digit * exp) % p;
        exp = exp * 10  % p;
        mods[current as usize] += 1;
        prev = current;
    }

    let mut answer = 0;
    for (i, cnt) in mods.iter().enumerate() {
        answer += cnt  * (cnt - 1)/ 2;
    }
    println!("{}", answer);
}