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;
typedef long long ll;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const int N=2e5+5;
const int mod=1e9+7;

ll n;

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	if(n==2) return cout<<1,0;
	ll ans=0;
	for(ll i=2;i*i<=n;++i){
		if(n%i==0){
			ll num=n;
			while(num%i==0){
				num/=i;
			}
			if(num%i==1) ++ans;
		}
	}	
	--n;
	for(ll i=2;i*i<=n;++i){
		if(n%i==0){
			if(n/i==i) ++ans;
			else ans+=2;
		}
	}
	ans+=2;
	cout<<ans;
	return 0;
}
