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 n,m,i,j,k,ans,cur,c;
	cin >> n;
	
	ans=0;
	for(i=1; i*i<=n-1; i++){
		if((n-1)%i==0){
			ans++;
			if(i*i<n-1){
				ans++;
			}
		}
	}
	ans--; // delete i==1
	
	for(i=2; i*i<=n; i++){
		if(n%i==0){
			k=n;
			while(k%i==0){
				k/=i;
			}
			if(k%i==1){
				ans++;
			}
			
			if(i*i<n){
				k=n;
				while(k%(n/i)==0){
					k/=(n/i);
				}
				if(k%(n/i)==1){
					ans++;
				}
			}
		}
	}
	ans++; // add i=n;
	cout << ans << endl;
	return 0;
}