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: Time Limit Exceeded)

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();
        scanner.nextLine();
        String s = scanner.nextLine().trim();
        System.out.println(numberOfDivisibleSubstrings(s, p));
    }

    public static long numberOfDivisibleSubstrings(String s, int p){
        long[] currentMods = new long[p];
        long[] previousMods = new long[p];
        long[] tmp;
        long counter = 0;
        for (char c : s.toCharArray()) {
            int digit = c-'0';
            for (int i = 0; i < p; i++) {
                currentMods[(i*10+digit)%p] += previousMods[i];
            }
            currentMods[digit%p]++;
            counter+=currentMods[0];
            tmp = currentMods;
            currentMods = previousMods;
            previousMods = tmp;
            Arrays.fill(currentMods,0);
        }
        return counter;
    }
}