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>
#include <iostream>
#include <algorithm>
#include <vector>


using namespace std;

int main(){
 long long int n;
 cin >> n;
  
 long long int ans = 0;
  
 long long int num = n;
  
 for (long long int i = 1; i*i <= n-1; i++) {
   
 if( (n-1) % i == 0){
   ans += 1;
   if((n-1)/i != i){
     ans += 1;
   }
  }
   
 }
  
  for(long long int i=2; i*i <= n;i++){
    num = n;
    if(n % i == 0){             
      while(num % i ==0){
        num /= i;
      }
      if(num % i == 1){
        ans++;
      }
    }
  }
  
  cout << ans << endl;
}