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 <vector>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
	long long n;
	cin>>n;
	int ans=0;
	for(long long i=1;i*i<=n-1;++i){
		if((n-1)%i==0){
			if(i!=(n-1)/i)
				ans+=2;
			else ans++;
		}
	}
	for(long long i=2;i*i<=n;++i){
		if(n%i==0){
			long long k=n;
			while(k%i==0){
				k/=i;
			}
			if(k%i==1) ans++;
		}
	}
	cout<<ans;
	return 0;
}