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;

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