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 PyPy3 (2.4.0) (judge verdict: Runtime Error)

import sys
input = sys.stdin.readline

n, p = [int(x) for x in input().split()]
s = input().rstrip()[::-1]
ans = 0
if p == 2:
    for i in range(n):
        if s[i] % 2 == 0:
            ans += (n - i + 1)
    print(ans)
    sys.exit()
if p == 5:
    for i in range(n):
        if s[i] == 0 or s[i] == 5:
            ans += (n - i + 1)
    print(ans)
    sys.exit()
t = [0, int(s[0]) % p]
mul = 10
for i in range(1, n):
    t.append((t[-1] + int(s[i])*mul) % p)
    mul *= 10
    mul %= p
from collections import Counter
c = Counter(t)
for i in c.keys():
    ans += c[i]*(c[i] - 1) // 2
print(ans)