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 ;
typedef long long ll ;
vector<ll> get(ll n)
{
  vector<ll> v ;
  for(ll i = 1; i * i <= n ;i ++)
   {
     if(n % i == 0)
      {
        v.push_back(i) ;
        if(i * i != n) v.push_back(n / i) ;
      }
   }
   return v ;
}
int main()
{
  ll n ;
  cin >> n ;
  auto v = get(n - 1) ;
  ll ans = v.size() - 1 ;
  v = get(n) ;
  for(ll q : v)
   {
     if(q == 1) continue ;
     ll z = n ;
     while(z % q == 0) z /= q ;
     if(z % q == 1) ans ++ ;
   }
   cout << ans << endl ;
  return 0 ;
}
