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 (3.8.2) (judge verdict: Wrong Answer)

from math import factorial
import collections

def permutations_count(n, r):
    ''' 順列 '''
    return factorial(n) // factorial(n - r)


def combinations_count(n, r):
    ''' 組み合わせ '''
    return factorial(n) // (factorial(n - r) * factorial(r))


n,p = map(int, input().split())
S = input()
su = 0

if p > 6 or p == 3:

    lis = []
    amari = 1
    lis.append(int(S[-1]) % p)

    for i in range(2, len(S)+1):
        amari = (amari * 10) % p
        lis.append((int(S[-1*i]) * amari + lis[i-2]) % p)

    c = collections.Counter(lis)

    for k in c:
        if c[k] > 1:
            su += combinations_count(c[k], 2)
    print(c)

    su += c[0]

    print(su)