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)

#include<bits/stdc++.h>
#define slowanddetermined ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define debug(x); cout<<#x<<"="<<x<<endl;
#define debug2(x,y); cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl;
#define int long long
using namespace std;



int fun(int n,int x){
    if(x==1) return 0;
    while(n%x==0){
        n/=x;
    }
    n=n%x;
    return n==1;
}

int32_t main()
{
   

    int n;
    cin>>n;
    set<int>st;
    for(int i=1;i*i<=(n-1);i++){
        if((n-1)%i==0){
            st.insert(i);
            st.insert((n-1)/i);
        }
    }
   
    for(int i=1;i*i<=n;i++){
        if(n%i==0){
            if(fun(n,i))st.insert(i);
            if(fun(n/i,i))st.insert(n/i);
        }
    }
    //c++;
     st.erase(1);
     cout<<st.size();


    return 0;
}