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(non_snake_case)]

use proconio::input;
use proconio::fastout;

fn powmod(mut b: usize, mut e: usize, m: usize) -> usize {
    let mut result = 1;
    while e > 0 {
        if e & 1 > 0 {
            result = (result * b) % m;
        }
        b = (b * b) % m;
        e >>= 1;
    }
    result
}

fn invmod(x: usize, p: usize) -> usize {
    powmod(x, p - 2, p)
}

#[fastout]
fn main() {
    input! {
        N: usize,
        P: usize,
        S: String,
    }
    let S = S.as_bytes();
    let mut result = 0usize;
    if 10 % P == 0 {
        for i in 0..N {
            let x = (S[i] - b'0') as usize;
            if x % P == 0 {
                result += i + 1;
            }
        }
    } else {
        let i10 = invmod(10, P);
        let mut dp = vec![0; P];
        let mut r = 0;
        for c in S {
            dp[r] += 1;
            let x = (c - b'0') as usize;
            r = ((r + P - x) * i10) % P;
            result += dp[r];
        }
    }
    println!("{}", result);
}