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: Accepted)

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 s = sc.next();
      long res = 0;
      if(p == 2 || p == 5) {
          for(int i = s.length() - 1; i >= 0; i--) {
            if(s.charAt(i) == '0' || (s.charAt(i) - '0') % p == 0)
                res += (i + 1);
          }
          System.out.println(res);
          return;
      }
      Map<Integer, Integer> prefixCnt = new HashMap<>();
      prefixCnt.put(0, 1);
      int fac = 1; 
      int cur = 0;
      
      for(int i = s.length() - 1; i >= 0; i--) {
        cur = (cur + (s.charAt(i) - '0') * fac) % p;
        fac = (fac * 10) % p;
        if(prefixCnt.containsKey(cur)) {
          res += prefixCnt.get(cur);
          prefixCnt.put(cur, prefixCnt.get(cur) + 1);
        } else
          prefixCnt.put(cur, 1);
      }
      System.out.println(res);
  }
}