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>
#define int long long
using namespace std;
int n,ans;
void find(int x)
{
	for(int i=1;i*i<=x;i++)
	if(x%i==0)
	{
		if(i!=1)
		ans++;
		if(i*i!=x)
		ans++;
	}
}
int pd(int x,int y)
{
	if(y==1)
	return 0;
	while(x%y==0)
	x/=y;
	if(x%y==1)
	return 1;
	return 0;
}
signed main()
{
	scanf("%lld",&n);
	find(n-1);
	for(int i=1;i*i<=n;i++)
	if(n%i==0)
	{
		ans+=pd(n,i);
		if(i*i!=n)
		ans+=pd(n,n/i);
	}
	printf("%lld",ans);
	return 0;
}