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: Accepted)

use proconio::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default, PartialEq, Eq, PartialOrd, Ord)]
struct Struct;

//#[proconio::fastout]
fn main() {
    input! {
        n: usize,
        p: usize,
        s: Bytes,
    }

    let mut ans = 0u64;
    if p == 2 || p == 5 {
        for i in 0..n {
            if (s[i] - b'0') as usize % p == 0 {
                ans += i as u64 + 1;
            }
        }
    } else {
        // hist[i] : s[j]以降を整数とみなしたとき値 % P == i となるjの値が何個あるか
        // 0<=j<=n
        let mut hist = vec![0; p];
        hist[0] = 1;
        let mut val = 0;
        let mut mul = 10;
        for i in (0..n).rev() {
            val = ((s[i] - b'0') as usize * mul + val) % p;
            mul = mul * 10 % p;
            hist[val] += 1;
        }
        for i in 0..p {
            let x = hist[i];
            if x > 0 {
                ans += x * (x - 1) / 2;
            }
        }
    }

    println!("{}", ans);
}