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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int check(int n, int k){
 int new_n = n / k;
 return new_n;
}

int main() {
  int n,N;
  cin >> n;
  N = n;
  
  int count = 1;
  bool flag = true;
  for (int i = 2; i < N; i++){
    n = N;
    flag = true;
  	while(flag){
        if ((n % i)==1){
          count = count + 1;
          flag = false;
        }
	    else if ((n % i)==0){
    	  n = check(n, i);
	    }
    	else{
          flag = false;
        }
    }
  }
  
  cout << count << endl;

  return 0;
}