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 <stdlib.h>
#include <string.h>
#include <math.h>

//#define MOD 10007
#define MOD 100003
//#define MOD 1000000007

typedef long long ll;

#define rep(i, n) for(i=0;i<(int)(n);i++)
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
#define bit(n) (1LL<<(n))
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return (a*b)/gcd(a,b);}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int res[MOD+2]={-1};

void rev_string(char str[])
{
    int i;
    int len = strlen(str);
    for (i = 0; i < len / 2; i++) {
        char temp = str[i];
        str[i] = str[len-i-1];
        str[len-i-1] = temp;
    }
}

int main(void){
  int n, p, i, j,result=0,tmp;
  char s[200001]={'\0'};
  
  scanf("%d %d", &n, &p);
  scanf("%s", s);
  
  for(i=0;i<n;i++){
    tmp=0;
    for(j=i;j<n;j++){
      tmp *= 10;
      tmp %= p;
      tmp += (int) (s[j]-'0');
      tmp %= p;
      
      if(tmp == 0){//equal mod 0
        result++;
      }
    }
  }
  printf("%d", result);
  
  return 0;
}