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;
bool test(long long a,long long b)
{
	while (a%b==0)
	a/=b;
	return (a%b==1);
}
int main()
{
	long long n;
	cin >> n;
	int ans=0;
	for (int i=2;1LL*i*i<=n-1;i++)
	{
		if ((n-1)%i==0)
		{
			ans++;
			ans+=(1LL*i*i!=n-1);
		}
	}
	for (int i=2;1LL*i*i<=n;i++)
	{
		if (n%i==0)
		{
			ans+=test(n,i);
			if (1LL*i*i!=n)
			ans+=test(n,n/i);
		}
	}
	cout << (ans+1+(n!=2));
}