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 <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
long long n,ans=1;
long long calc1(long long x)
{
	long long now=n;
	while (now%x==0) now/=x;
	if (now%x==1) return 1;
return 0;
}
int main()
{
	scanf("%lld",&n);
	for (long long i=2;i*i<=n;i++)
	if (n%i==0) 
	{
		ans+=calc1(i);
		if (i*i!=n) ans+=calc1(n/i);
	}
	//printf("lalal %lld\n",ans);
	if (n>2) ans++;
	for (long long i=2;i*i<=n-1;i++)
	if ((n-1)%i==0)
	{
		if (n%i!=0) ans++;
		if (n%((n-1)/i)!=0 && (i*i!=(n-1))) ans++;
	}
	printf("%lld\n",ans);
return 0;
}