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: Accepted)

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

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int P = sc.nextInt();
		String S = sc.next();
		sc.close();

		if ( P == 2 || P == 5 ) {
			long ans = 0;
			for ( int i = N - 1 ; i >= 0 ; i-- ) {
				int digit = S.charAt(i) - '0';
				if ( digit % P == 0 ) {
					ans += i + 1;
				}
			}
			System.out.println(ans);
		} else {
			int[] mods = new int[P];
			mods[0] = 1;
			int D = 1;
			int last = 0;
			long ans = 0;
			for ( int i = N - 1 ; i >= 0 ; i-- ) {
				int digit = S.charAt(i) - '0';
				last = (D * digit + last) % P;
				ans += mods[last];
				mods[last]++;
				D = (D * 10) % P;
			}
			System.out.println(ans);
		}
	}
}