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)

# coding: utf-8
import numpy as np
N,P = list(map(int, input().split()))
S = '0'+input()

#N=200000
#P=9893
#S= '0'+'1234567890' * 20000

if (P==2)|(P==5):
    sum=0
    for i in range(N, 0, -1):
        if int(S[i]) % P == 0:
            sum+= i
    print(sum)


else:
    dp1 = np.zeros(N + 1)
    dp2 = np.zeros(N + 1)
    dp3 = np.zeros(P)
    dp1[N] = int(S[N]) % P
    dp3[int(dp1[N])] += 1
    k = 1
    for i in range(N - 1, 0, -1):
        dp1[i] = (dp1[i + 1] + (int(S[i]) * 10 * k)) % P
        dp3[int(dp1[i])] += 1
        k = (k * 10) % P

    sum = 0

    for i in range(P):
        if i == 0:
            sum += (dp3[i] * (dp3[i] - 1)) // 2 + dp3[i]
            continue
        if dp3[i] == 0: continue
        sum += (dp3[i] * (dp3[i] - 1)) // 2
    print(dp1)
    print(dp3)
    print(int(sum))