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
//F 
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
	ll n,ans = 2;
	cin >> n;
	for(ll i = 2;i*i <= n-1;i++){
		if((n-1)%i == 0){
			ans += 2;
			if((n-1)/i == i)ans--;
		}
	}
	for(ll k = 2;k*k <= n;k++){
		if(n%k)continue;
		ll u = n;
		while(u%k == 0)u /= k;
		u %= k;
		if(u%k == 1 or u == 1)ans++; 
	}
	if(n == 2)ans--; 
	cout << ans;
	return 0;
}
