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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <assert.h>
#include <stdio.h>
 
int n, p, i, t, y = 1, d[10000] = {1};
long long r;
char s[200002];
 
int main() {
  assert(scanf("%d%d%s", &n, &p, s + 1) == 3);
  if (10 % p == 0) {
    while(i++<n) {
      if((s[i] - '0') % p == 0)
        r += i;
    }
  }
  else {
    for(; n; n--) {
      t = (t + (s[n] - '0') * y) % p;
      d[t]++;
      y = y * 10 % p;
    }
    for(i = 0; i < p; i++)
      r += (long long) d[i] * (d[i] - 1);
    r /= 2;
  }
  printf("%lld\n", r);
  return 0;
}