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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int p = scanner.nextInt();
		String s = scanner.next();
		long count;

		if (p == 2 || p == 5) {
			count = 0;
			for (int i = 0; i < n; i++) {
				if ((s.charAt(i) - 48) % p == 0) {
					count += i + 1;
				}
			}
		} else {

			int[] place = new int[p - 1];//10^a mod p=10^(a%(p-1)) mod p
			place[0] = 1;
			for (int i = 1; i < p - 1; i++) {
				place[i] = place[i - 1] * 10 % p;
			}

			int[] r = new int[p];
			r[0] = 1;
			int remain = 0;
			for (int i = 0; i < n; i++) {
				remain += (s.charAt(n - 1 - i) - 48) * place[i % (p - 1)];
				remain %= p;
				r[remain]++;
			}

			count = Arrays.stream(r).parallel().map(i -> (int)((long)i * (i - 1) / 2)).sum();

		}

		System.out.println(count);

	}

}