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 <iostream>
#include <set>
#include <vector>
using namespace std;

typedef long long ll;
set<ll> ans;

int main() {
  ll n;
  cin >> n;

  for(ll i = 2; i * i <= n; i++) {
    if(n % i == 0) {
      ll tmp = n;
      while(tmp >= i) {
	if(tmp % i == 0) {
	  tmp /= i;
	}
	else tmp %= i;

	if(tmp == 1) {
	  ans.insert(i);
	  break;
	}
      }
    }
  }
  ans.insert(n);
  if(n > 2) ans.insert(n-1);
  n--;
  for(ll i = 2; i * i <= n; i++) {
    if(n % i == 0) {
      ans.insert(i);
      ans.insert(n / i);
    }
  }
  cout << ans.size() << endl;
}
