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>
#define int long long
using namespace std;

long long n, cnt, a[200005], ans;

signed main() {
	cin>>n;
	for(int i = 2; i <= sqrt(n); i++) 
	     if(n%i == 0) {
	     	a[++cnt] = i;
	     	if(i*i != n) a[++cnt] = n/i;
	     }
	for(int i = 1; i <= sqrt(n - 1); i++)  {
		if((n - 1)%i == 0) {
			ans += 2;
			if(i*i == n - 1) ans--;
		}
	}
	for(int i = 1; i <= cnt; i++) {
		int t = 0;
		long long s = n, res = 1;
		while(s%a[i] == 0) {
			t++;
			s /= a[i];
			res *= a[i];
		}
		
		if((n - res)%(res*a[i]) == 0) ans++;
	}
	cout<<ans<<endl;
	
}