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() {
  
  long n;
  cin >> n;
  unordered_set<long> factors({n});
  for (long k = 2; k * k <= n; ++k) {
    if (n % k == 0) {
      long x(n);
      while (x % k == 0) x /= k;
      if (x % k == 1) factors.insert(k);
    }
  }
  for (long k = 1; k * k <= n - 1; ++k) {
    if ((n - 1) % k == 0) {
      factors.insert(k);
      factors.insert((n - 1) / k);
    }
  }
  cout << factors.size() - 1 << '\n';
  return 0;
}