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 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int main()
{
    long long n;
    cin>>n;
    long long ttl=0ll;
    for(long long i=1ll;i*i<=n-1ll;i++){
        if((n-1ll)%i==0ll&&(n)%i!=0){
            if(i!=1ll)
                ttl++;
            if(n%((n-1)/i)!=0)
            ttl++;
        }
    }
    for(long long i=1ll;i*i<=n;i++){
        if(n%i==0ll){
            if(i!=1ll){
                long long tmn=n;
                while(tmn%i==0)
                    tmn/=i;
                if((tmn-1)%i==0)
                    ttl++;
            }
            if(i==(n/i))
                continue;
            long long tmn=n;
            while(tmn%(n/i)==0){
                tmn/=(n/i);
            }
            if((tmn-1)%(n/i)==0)
                ttl++;
        }
    }
    cout<<ttl<<'\n';
    return 0;
}