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;
#define ll long long
ll n,m,i,j,a[100001],sum=1;
int main()
{
	cin>>n;
	for(i=1;i*i<n;i++)
	{
		ll k=n-1;
		if(k%i==0)
		{
			if(i*i!=k)
			{
				sum++;
			}
			if(i!=1)
			{
				sum++;
			}
		}
	}
	for(i=2;i*i<=n;i++)
	{
		ll k=n;
		while(k%i==0)
		{
			k/=i;
		}
		if(k!=n&&k%i==1)
		{
			sum++;
		}
	}
	cout<<sum<<"\n";
	return 0;
}