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;
int main(void){
    // Your code here!
 long long int i,c,k,m,n;
 c=0;
 cin >> n;
 k=n-1;
 for(i=1;i<=sqrt(k);i++)
 {
  if(k%i==0)
  {
   c++;
   if(i!=(k/i))
   c++;
  }
 }
 for(i=2;i<=sqrt(n);i++)
 {
  if(n%i==0)
  {
   m=n;
   while(m%i==0)
   m/=i;
   if(m%i==1)
   c++;
   if(i!=(n/i))
   {
    m=n;
    while(m%(n/i)==0)
    m/=(n/i);
    if(m%(n/i)==1)
    c++;
   }
  }
 }
 cout << c;
 return 0;
}