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: Time Limit Exceeded)

N, P = [int(i) for i in input().split()]
S = input()

if P == 2 or P == 5:
#if False:
  count = 0
  for n in range(N):
    if int(S[n]) % P==0:
      count += (n + 1)
  print(count)
else:    
  Pa = [0 for _ in range(P)]
  number = 0
  for i in range(N, 0, -1):
#    ui = int(S[i-1:])
    number = number + int(S[i-1])*10**(N-i)
    mp = number % P
    Pa[mp] += 1
  count = 0
  if Pa[0] > 0:
    count += Pa[0]
  for p in range( P):
    pmod = Pa[p]
    if pmod > 0:
      count += (pmod * (pmod - 1))//2
  print(count)