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 int 
using namespace std;

// Author: Ashish

int32_t main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin>>n;
  if(n==2) {cout<<"1"; return 0;}
  set<int> answer;
  n=n-1;
  answer.insert(2);
  for(int i=1;i*i<=n;i++)
  {
  	if(n%i==0) answer.insert(i),answer.insert(n/i);
  }
  n=n+1;
  for(int i=2;i*i<=n;i++)
  {
  	if(n%i==0)
  	{
  		int temp=n;
  		while(temp%i==0)
  			temp/=i;
  		if(temp%i==1) answer.insert(i);
  	}
  }
  answer.insert(n);
  cout<<answer.size()-1;
 	return 0;
}