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.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main {

    public static void main(final String[] args) throws Exception {
        final Scanner sc = new Scanner(System.in);
        long N;
        N = sc.nextLong();
        String Pstr;
        Pstr = sc.next();
        String S;
        S = sc.next();

        final Solver solver = new Solver();
        solver.solve(N, Pstr, S);
    }
}

class Solver {
    public Map<String, Boolean> dividable = new HashMap<>();

    public void solve(long N, String Pstr, String S) {
        BigInteger P = new BigInteger(Pstr);

        long ans = 0;
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j <= N; j++) {
                String sub = S.substring(i, j);
                if (isDividable(sub, P)) ans++;
            }
        }
        System.out.println(ans);
    }

    public boolean isDividable(String d, BigInteger P) {
        if (!dividable.containsKey(d)) {
            BigInteger b = new BigInteger(d);
            dividable.put(d, b.remainder(P).equals(BigInteger.ZERO));
        }

        return dividable.get(d);
    }
}