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::*;

fn main() {
    let mut s: String = String::new();
    std::io::stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let p: usize = itr.next().unwrap().parse().unwrap();
    let mut s: Vec<usize> = itr
        .next()
        .unwrap()
        .chars()
        .map(|c| c as usize - '0' as usize)
        .collect();

    let mut ans = 0;
    if p == 2 {
        for i in 0..n {
            if s[i] % 2 == 0 {
                ans += i + 1;
            }
        }
    } else if p == 5 {
        for i in 0..n {
            if s[i] % 5 == 0 {
                ans += i + 1;
            }
        }
    } else {
        s.reverse();
        let mut memo: Vec<usize> = vec![0; p + 1];
        let mut now = 0;
        memo[0] = 1;
        for i in 0..n {
            now = (now * 10 + s[i]) % p;
            memo[now] += 1;
        }
        for i in 0..p + 1 {
            if memo[i] > 0 {
                ans += memo[i] * (memo[i] - 1) / 2;
            }
        }
    }
    println!("{}", ans);
}