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;
int main(){
	ll n,ans=2;//初始值包括n和n-1两种. 
	cin>>n;
	if(n==2) ans=1;//特判 
	for(ll k=2,i=n;k*k<=n;k++,i=n){ 
		while(i>=k) i%k?i%=k:i/=k;//这里是计算[2,sqrt[n]的情况. 
		if(i==1) ans++;
		if((n-1)%k==0&&k*k!=(n-1)) ans++;//这里是计算 [sqrt(n),n]n-1的因数情况，这里k*k!=(n-1)是去重 
	}
	cout<<ans<<endl;
	return 0;
}
