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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
  long n;
  scanf("%ld", &n);
  
  long k = 2;
  long cnt = 0;
  long tmp;
  
  while (k < n)
  {
    tmp = n;
    if (tmp == 1)
    {
      cnt++;
    }
    else if (tmp % k == 0)
    {
      tmp = tmp / k;
    }
    else
    {
      tmp = tmp - k;
    }
    k++;
  }
  
  printf("%ld", cnt);
  return 0;
}