Problem:
Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program division_or_substraction
  implicit none
  integer(8) :: n, r, k, x = 2_8
  read(*,*) n
  if (n == 2_8) then
    write(*,'(i0)') 1
    stop
  end if
  r = sqrt(real(n,8)+0.5d0)
  do k = 2_8, r
    if (ok(n,k)) then
      x = x+1_8
    else
      if (mod(n-1_8,k) == 0_8) x = x+1_8
      if (k*k /= n-1_8) x = x+1_8
    end if
  end do
  write(*,'(i0)') x
contains
  logical function ok(n,k)
    integer(8), intent(in) :: n, k
    integer(8) :: m
    m = n
    do while (m >= k)
      if (mod(m,k) == 0) then
        m = m/k
      else
        m = mod(m,k)
      end if
    end do
    ok = m == 1
  end
end program division_or_substraction