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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;

ll n;

inline bool ok(ll x)
{
	ll tmp=n;
	while(tmp && tmp%x==0) tmp/=x;
	return tmp%x==1;
}

int main(void)
{
	scanf("%lld",&n);
	
	ll ans=0;
	for(int i=1; (ll)i*i<=n; ++i)
		if(n%i==0)
		{
			if(i>1) ans += ok(i);
			if((ll)i*i!=n) ans += ok(n/i);
		}
	
	--n;
	for(int i=1; (ll)i*i<=n; ++i)
		if(n%i==0)
		{
			if(i>1) ++ans;
			if((ll)i*i!=n) ++ans;
		}
	printf("%lld",ans);
	return 0;
}