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>
using namespace std;
int main()
{
    long long n;
    cin>>n;
    long long m=n-1;
    long long ans1=0;
    for(long long i=2;i<=sqrt(m);i++)
    {
        if(m%i==0)
        {
            ans1+=2;
        }
        if(i*i==m) ans1--;
    }
    for(long long i=2;i<=sqrt(n);i++)
    {
        if(n%i==0)
        {
            long long p=n;
            while(p%i==0&&p)
            {
                p/=i;
            }
            if((p-1)%i==0)  ans1++;
        }
    }
    ans1++;
    cout<<ans1+1;
}