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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int main(void) {
    int N, P;
    int ans=0;
    char S[200000];
    int Snum[200000];
    int dp2[200000];
    int tmp;
    scanf("%d%d%s", &N, &P, S);
    for (int i = 0; i<N; i++){
        // Snum[i]にはi桁目の数が入る
        Snum[i] = (S[N-1-i] - '0')%P;
    }

    // dp2[0] = Snum[0] == 0 ? 1 : 0;
    for(int i=0;i<N; i++){
        tmp = 0;

        for (int k=i; k>=0; k--){
            tmp = (tmp*10+Snum[k])%P;
            if (tmp == 0) {
                dp2[i] = k == 0 ? 1 : dp2[k-1] + 1;
                break;
            }
        }
    }
    ans = 0;
    for (int i = 0; i < N; i++){
        ans += dp2[i];
    }
     
    printf("%d\n", ans);
 
    return 0;
}