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: Accepted)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let p: i32 = s.next().unwrap().parse().unwrap();
	let s: Vec<_> = s.next().unwrap().bytes().rev().collect();
	let mut ans = 0;
	if p == 2 || p == 5 {
		for (i, c) in s.iter().enumerate() {
			let d = (c - b'0') as i32;
			if d % p == 0 {
				ans += n - i;
			}
		}
	} else {
		let mut pos = vec![vec![]; p as usize];
		let mut base = 1;
		let mut r = 0;
		for (i, c) in s.iter().enumerate() {
			r = (r + base * (c - b'0') as i32) % p;
			pos[r as usize].push(i);
			base = (base * 10) % p;
		}
		let mut base = 1;
		let mut r = 0;
		for (i, c) in s.iter().enumerate() {
			ans += pos[r as usize].len() - pos[r as usize].lower_bound(&i);
			r = (r + base * (c - b'0') as i32) % p;
			base = (base * 10) % p;
		}
	}
	println!("{}", ans);
}

trait BinarySearch<'a, T: 'a> {
	fn lower_bound(&self, x: &T) -> usize;
}

impl<'a, T: 'a + Ord> BinarySearch<'a, T> for [T] {
	fn lower_bound(&self, x: &T) -> usize {
		let mut first = 0;
		let mut len = self.len() - first;
		while len > 0 {
			let half = len >> 1;
			let mid = first + half;
			use std::cmp::Ordering;
			match self[mid].cmp(x) {
				Ordering::Less => {
					first = mid + 1;
					len = len - half - 1;
				}
				Ordering::Equal | Ordering::Greater => {
					len = half;
				}
			}
		}
		first
	}
}