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.

Examples

Input

6


Output

3


Input

3141


Output

13


Input

314159265358


Output

9
#include "bits/stdc++.h"
using namespace std;

long long n;

int div_count(long long n)
{
	int ans=0;
	bool prime=true;
	for(int i=1;1ll*i*i<=n;i++)
	{
		if(n%i==0)
		{
			ans+=2;
			if(1ll*i*i==n) ans--;
			prime=false;
		}
	}
	return ans+prime-1;
}

bool ok(long long i)
{
	long long m=n;
	while(m%i==0) m/=i;
	return m%i==1;
}

int main()
{
	cin>>n;

	int ans=div_count(n-1);
	for(int i=2;1ll*i*i<=n;i++)
	{
		if(n%i) continue;
		ans+=ok(i);
		if(1ll*i*i!=n) ans+=ok(n/i);
	}
	cout<<ans+1;
}