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 ;

const int MAX = 1e5 + 10 ;

int arr[MAX] ;
long long n ;

long long ans = 0 ;

void solve(long long k)
{
	long long a = n ;
	if(k == 1)
	{
		if(a == 2)
			ans++ ;
		return ;
	}
	while(a % k == 0)
		a /= k ;
	if(a % k == 1)
		ans++ ;
	return ;
}

void factorize(long long x)
{
	if(x == n-1 && x != 1)
		solve(x) ;
	if(x == n)
		solve(1) , solve(x) ;
	for(long long i = 2ll ; i * i <= x ; ++i)
	{
		if(x % i == 0)
		{
			solve(i) ;
			if(x/i != i)
				solve(x/i) ;
		}
	}
	return ;
}

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	cin>>n ;
	factorize(n-1) ;
	factorize(n) ;
	return cout<<ans<<"\n" , 0 ;
}