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;


int main()
{
	long long n;
	long long ans=0,s=1,c;
	cin>>n;
	if(n==2) {
		printf("1");
		return 0;
	}
	long long maxx=sqrt(n);
	long long t=n-1;
	for(int i=2;i<=maxx;++i) {
		if(t%i==0) {
			c=1;
			while(t%i==0) t/=i,c++;
			s*=c;
		}
	}
	if(t!=1) s*=2;
	for(int i=2;i<=maxx;++i) {
		if(n%i==0) {
			long long t=n;
			while(t%i==0) t/=i;
			if(t%i==1) ans++;
		}
	}
	cout<<ans+s;
}