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 <stdio.h>
#include <stdlib.h>

int cmpnum(const void * n1, const void * n2) {
	if (*(long *)n1 > *(long *)n2)
	{
		return 1;
	}
	else if (*(long *)n1 < *(long *)n2)
	{
		return -1;
	}
	else
	{
		return 0;
	}
}

int main(void) {

  long n,p;
  scanf("%ld %ld", &n, &p);
  char s[n+1];
  scanf("%s", s);
  long sum = 0;
  long num;
  if (p == 2) {
    for (long i = 0; i < n; i++) {
      num = s[i]-'0';
      if (num%2 == 0) {
        sum += i+1;
      }
    }
  } else if (p == 5) {
    for (long i = 0; i < n; i++) {
      num = s[i]-'0';
      if (num%5 == 0) {
        sum += i+1;
      }
    }
  } else {
    long mod_list[n];
    mod_list[0] = s[n-1]-'0';
    mod_list[0] %= p;
    long unit = 1;
    for (long i = 1; i < n; i++) {
      num = s[n-1-i]-'0';
      unit = (unit*10)%p;
      mod_list[i] = mod_list[i-1];
      mod_list[i] += num*unit;
      mod_list[i] %= p;
    }
    qsort(mod_list, n, sizeof(long), cmpnum);
    long count = 0;
    for (long i = 0; i < n-1; i++) {
      count++;
      if (mod_list[i] != mod_list[i+1]) {
        sum += count*(count-1)/2;
        count = 0;
      }
    }
    count++;
    sum += count*(count-1)/2;
    count = 0;
    for (long i = 0; i < n; i++) {
      if (mod_list[i] == 0) {
        count++;
      } else {
        break;
      }
    }
    sum += count;
  }
  printf("%ld\n", sum);

  return 0;
}