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;
long long n;
int ans=0;
void calc(long long x){
	long long a=n;
	while(a%x==0){
		a/=x;
	}
	if(a%x==1){
		ans++;
	}
}
int main(){
	cin>>n;
	for(int i=1;(long long)i*i<n;++i){
		if((n-1)%i==0){
			if(i>1){
				ans++;
			}
			if((n-1)/i!=i){
				ans++;
			}
		}
	}
	for(int i=1;(long long)i*i<=n;++i){
		if(n%i==0){
			if(i>1){
				calc(i);
			}
			if(i!=n/i){
				calc(n/i);
			}
		}
	}
	cout<<ans<<'\n';
	return 0;
}