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.*;

public class Main {

    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    long p = sc.nextInt();

    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

        char[] s = sc.next().toCharArray();

        long ans = 0;

        long sumMod = 0;

        Map<Long, Long> map = new HashMap<>();

        for (int i=n-1; 0<=i; i--) {
            long tmp = s[i] - '0';
            tmp *= modPow(10, n-1-i);
            tmp %= p;
            sumMod += tmp;
            sumMod %= p;
            if (map.containsKey(sumMod)) {
                map.put(sumMod, map.get(sumMod)+(long)1);
            } else {
                map.put(sumMod, (long)1);
            }
        }

//        System.out.println(map);

        for (Long t: map.values()) {
            ans += t * (t - 1) / 2;
        }

        if (map.get((long)0) == null) {
            System.out.println(ans);
        } else {
            System.out.println(ans + map.get((long)0));
        }


    }

    long modPow(long x, long n) {
        long sum = 1;
        while (n > 0) {
            if ( (n & 1) == 1 ) {
                sum *= x;
                sum %= p;
            }
            x *= x;
            x %= p;
            n >>= 1;
        }
        return sum % p;
    }
}