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: Time Limit Exceeded)

use std::io::{self, BufRead};

fn main() {
  let stdin = io::stdin();
  let lines: Vec<String> = stdin.lock().lines().map(|x| x.unwrap()).collect();
  let xs: Vec<i32> = lines[0].split(' ').map(|x| x.parse().unwrap()).collect();
  let p = xs[1];
  let s = &lines[1];
  let bs = s.as_bytes();
  let n = s.len();
  let mut count = 0;

  for i in 0..n {
    let mut m0 = 0;
    for j in i..n {
      let x = (bs[j] as i32) - 48;
      let m1 = (m0 * 10 + x) % p;
      if m1 == 0 {
        count += 1;
      }
      m0 = m1;
    }
  }

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