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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(void)
{
  long long int N,M1,M2,K;
  int c=2;

  scanf("%lld",&N);

  if(N==2)
  {
    printf("1\n");
    return 0;
  }

  for(K=2;K<sqrt(N)+1;K++)
  {
    M1=N;
    M2=N;
    if(K*K==N-1) c++;
    else if((N-1)%K==0) c+=2;
    if(K*K==N) c++;
    else if(N%K==0)
    {
      while(M1%K==0) M1/=K;
      while(M2%(N/K)==0) M2/=N/K;
      if(M1%K==1) c++;
      if(M2%K==1) c++;
    }
  }

  printf("%d\n",c);

  return 0;
}