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<iostream>
#include<cmath>
using namespace std;
int main()
{
  long long n,m,t,i,sq,res=0;
  cin >> n;
  if(n==2){
    cout << "1" << endl;
    return 0;
  }
  sq=(long long)floor(sqrt(n)+1);
  for(i=2;i<=sq;i++){
    m=n;
    while(m%i==0)
      m/=i;
    if(m%i==1)
      res++;
    if(n%i==1){
      t=(n-1)/i;
      if(t>sq)
        res++;
    }
  }
  if(n-1>sq) res++;
  if(n>sq) res++;
  cout << res << endl;
  return 0;
}