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: Wrong Answer)

def ext_gcd(a, b, x, y):
    if b == 0:
        x[0] = 1
        y[0] = 0
        return a
    
    d = ext_gcd(b, a % b, y, x)
    y[0] -= a // b * x[0]

    return d

def get_inv(n, p):
    x = [1]
    y = [1]
    ext_gcd(n, p, x, y)
    return x[0] % p

n, p = map(int, input().split())
s = list(input())

for i in range(n):
    s[i] = int(s[i]) % p

ls = [0]
p10 = [1]

for i in range(n):
    ls.append((ls[i] * 10 + s[i]) % p)
    p10.append(p10[i] * 10 % p)

mp = [0 for i in range(p)]
ans = 0

for i in range(n + 1):
    tmp = get_inv(p10[i], p)
    ans += mp[ls[i] * tmp % p]
    mp[ls[i] * tmp % p] += 1

print(ans)