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: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
int my_atoi(char *, int);
int main(void)
{
  int n, p, i, j, temp_num, left_end_pointer, pointer, right_end_pointer, counter_of_divisible;
  char s[200001];

  scanf("%d %d", &n, &p);            // 標準入力からの値取得
  scanf("%s", s);                    // 標準入力からの値取得

  counter_of_divisible = 0;
  for (left_end_pointer = 0; left_end_pointer < n; left_end_pointer++) {
    for (right_end_pointer = left_end_pointer; right_end_pointer < n; right_end_pointer++) {
      temp_num = 0;
      pointer = left_end_pointer;
      for (i = left_end_pointer; i <= right_end_pointer; i++) {
        temp_num *= 10;
        if (i == pointer + 7 || i == right_end_pointer) {
          temp_num += my_atoi(&s[pointer], i - pointer + 1);
          temp_num %= p;
          pointer = i + 1;
        }
      }
      if (temp_num == 0) {
        counter_of_divisible++;
      }
    }
  }

  printf("%d\n", counter_of_divisible);   // 出力
  return 0;
}

int my_atoi(char *s, int i) {
  char temp = *(s + i);
  *(s + i) = '\0';
  int ret = atoi(s);
  *(s + i) = temp;
  return ret;
}