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: Wrong Answer)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
int main(){
    int cnt=0;
    int64_t n,sum=0,k,p;
    cin>>n;
    k=n;
    p=n;
    for(int64_t i=2;i<=sqrt(n);i++){
       while(n%i==0){
           n=n/i;
           while(k%n==0){
                k=k/n;
                if(k%n!=0)cnt--;    
           }
           if(n%i==1)cnt+=2;
       }
    }
    for(int64_t i=1;i<=sqrt(p-1);i++){
       if((p-1)%i==0){
           cnt+=2;
       }
    }
    sum=cnt;
    cout<<sum;
}