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: Time Limit Exceeded)

#![allow(non_snake_case)]

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

#[fastout]
fn main() {
    input! {
        N: usize,
        P: usize,
        S: String,
    }
    let mut result = 0;
    let mut dp = vec![0u64; P];
    let mut r = 0;
    for c in S.bytes() {
        let x = (c - b'0') as usize;
        let mut dp_next = vec![0u64; P];
        for r in 0..P {
            dp_next[(10 * r + x) % P] += dp[r];
        }
        dp_next[x % P] += 1;
        dp = dp_next;
        result += dp[0];
    }
    println!("{}", result);
}