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 int n,l,t;
	cin>>n;
	set<long long int> s;
	for(long long int i=2;i*i<=n-1;i++){
		if((n-1)%i==0){
			s.emplace(i);
			s.emplace((n-1)/i);
		}
	}
	for(long long int i=2;i*i<=n;i++){
		if((n%i)==0){
			t=n;
			while(t>=i){
				if(t%i==0)
				t=t/i;
				else t=t%i;
			}
			if(t==1)
			s.emplace(i);
			t=n;l=n/i;
			while(t>=l){
				if(t%l==0)
				t=t/l;
				else t=t%l;
			}
			if(t==1)
			s.emplace(l);
		}
	}
	s.emplace(n);
	if(n>2)
	s.emplace(n-1);
	cout<<s.size()<<endl;
}