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.*;

public class Main {

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

        for (int i = 0; i < n; i++) {
            s[i] = str.charAt(i) - '0';
        }

        int[][] dp = new int[2][p];

        for (int j = 0; j < p; j++) {
            dp[0][j] = 0;
            if ((s[0] % p) == j) {
                dp[0][j] = 1;
            }
        }

        long ans = dp[0][0];
        for (int i = 1; i < n; i++) {
            int next = i%2;
            int prev = 1-next;
            for(int j = 0; j < p; j++){
                dp[next][j] = 0;
            }
            for (int j = 0; j < p; j++) {
                dp[next][(j * 10 + s[i]) % p] += dp[prev][j];
                if ((s[i] % p) == j) {
                    dp[next][j] += 1;
                }
            }
            ans += dp[next][0];
        }

        System.out.println(ans);
    }
}