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<stdlib.h>
#include<math.h>
int main(){
    int n,p;
    scanf("%d %d",&n,&p);
    char *s;
    s=(char*)malloc(sizeof(char)*(n+1));
    scanf("%s",s);
    int ans=0;
    if (p==2||p==5)
    {
        for (int i = 0; i < n; i++)
        {
            if (((int)(s[n-1-i]-'0'))%p==0)
            {
                ans+=n-i;
            }
            
        }
        
    }else{
        int *mod;
        mod=(int*)malloc(sizeof(int)*p);
        for (int i = 0; i < p; i++)mod[i]=0;
        int amari=0;
        for (int i = 0; i < n; i++)
        {
            int u=((int)(s[n-1-i]-'0'));
            for (int j = 0; j < i; j++)
            {
                u*=10;
                u%=p;
            }
            
            amari=(amari+u)%p;
            mod[amari]++;
        }
        ans+=(mod[0]*(mod[0]+1))/2;
        for (int i = 1; i < p; i++)
        {
            ans+=(mod[i]*(mod[i]-1))/2;
        }        
    }
    printf("%d\n",ans);
    return 0;
}