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;

bool aaa(long a,long b) {
  long aa=a;
  while(aa%b==0) aa=aa/b;
  if(aa%b==1) return true;
  else return false;
}
int main() {
  long n,ans=0;
  cin >> n;
  for(long i=1;i*i<=n-1;i++) {
    if((n-1)%i==0) {
      if(i!=1) ans++;
      if(i*i!=n-1) ans++;
    } 
  }
  for(long i=1;i*i<=n;i++) {
    if(n%i==0) {
      if(i!=1) if(aaa(n,i)) ans++;
      if(i*i!=n) if(aaa(n,n/i)) ans++;
    }
  }
  cout << ans << endl;
}