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;

int n;

bool ok(int x) {
	if(x == 1) return false;
	int tt = n;
	while(tt % x == 0) tt /= x;
	return tt % x == 1;
}

signed main() 
{
	cin >> n;	
	int ans = 0;
	for(int i = 1 ; i * i <= n ; i++ ) {
		if(n % i == 0)  {
			ans += ok(i);
			if(i*i != n) ans += ok(n/i);
		}
	}
	int nn = n - 1;
	for(int i = 1 ; i * i <= nn ; i++ ) {
		if(nn % i == 0)  {
			ans += ok(i);
			if(i*i != nn) ans += ok(nn/i);
		}
	}
	cout << ans << endl;
}