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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // int64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

int get_str(char *str, int size)
{
    char format[100];
    if (size == 0)
        return 0;
    snprintf(format, sizeof(format), "%%%ds", size);
    scanf(format, str);
    return 0;
}

#define NUM_MAX 200000
#define PRIME_MAX 10000

int main(void) {
    int num = get_int();
    int divisor = get_int(); // assume that it's prime
    static char str[NUM_MAX];
    static int mod_map[PRIME_MAX];
    get_str(str, num);
    
    int i;
    int sum = 0;
    for(i = num-1; i >= 0; i--) {
        int d = str[i] - '0';
        sum = (sum*10+d)%divisor;
        mod_map[sum]++;
    }
    int64_t ans = mod_map[0];
    int offset = 0;
    for(i = num-1; i >= 1; i--) {
        int d = str[i] - '0';
        offset = ((offset*10)+d)%divisor;
        mod_map[offset]--;
        ans += mod_map[offset];
    }
    printf("%lld\n", ans);
    return 0;
}