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 Python (3.4.3) (judge verdict: Accepted)

def main():
    N,p = map(int, input().split())
    S = input()
    S = S[::-1]

    if p==2 or p==5:
        ans = 0
        for i,s in enumerate(S):
            if int(s)%p==0:
                ans += N-i
        print(ans)
    else:
        # modpを作成
        modp_counter = [0] * p
        modp = 0
        modp_counter[0] = 1 # 割り切れるやつが入る場所の初期値を1にしておかないと、1つ割り切れるやつがきたときにpair_count+=0となってしまう(pair_count+=1になるべきなのに)
        pair_count = 0
        d = 1
        for s in S:
            modp = (int(s)*d%p+modp)%p
            pair_count += modp_counter[modp]
            modp_counter[modp] += 1
            d *= 10
            d %= p # 計算量削減
        print(pair_count)

main()