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.BufferedReader;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int p = Integer.parseInt(sa[1]);
		char[] s = br.readLine().toCharArray();
		br.close();

		long ans = 0;
		Map<Integer, Integer> map = new HashMap<>();
		for (int i = 0; i < n; i++) {
			if (map.containsKey(0)) {
				map.put(0, map.get(0) + 1);
			} else {
				map.put(0, 1);
			}
			Map<Integer, Integer> work = new HashMap<>();
			for (Integer key : map.keySet()) {
				int next = key * 10 + (s[i] - '0');
				next %= p;
				int num = map.get(key);
				if (next == 0) {
					ans += num;
				}
				if (work.containsKey(next)) {
					work.put(next, work.get(next) + num);
				} else {
					work.put(next, num);
				}
			}
			map = work;
		}
		System.out.println(ans);
	}
}