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 Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program divisible_substring
  implicit none
  integer :: n, p, i, j, d, k = 0, cnt(0:10000) = 0, hash = 0, pow = 1
  character(200000) :: s
  integer(8) :: x = 0
  read(*,*) n, p
  read(*,*) s
  if (p == 2 .or. p == 5) then
    do i = 1, n
      d = ichar(s(i:i)) - 48
      if (mod(d, p) == 0) then
        x = x + i
      end if
    end do
    write(*,'(i0)') x
    stop
  end if
  cnt(0) = 1
  do i = n, 1, -1
    d = ichar(s(i:i)) - 48
    hash = mod(d * pow + hash, p)
    x = x + cnt(hash)
    cnt(hash) = cnt(hash) + 1
    pow = mod(10 * pow, p)
  end do
  write(*,'(i0)') x
end program divisible_substring