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;
using i64 = long long;
#define endl "\n"

int main()
{
  i64 N;
  cin >> N;
  set<i64> ans;
  ans.insert(2);
  ans.insert(N);
  if (2 < N)
    ans.insert(N - 1);
  for (i64 i = 2; i * i <= N; i++)
  {
    if ((N - 1) % i == 0)
    {
      ans.insert(i);
      ans.insert((N - 1) / i);
    }
    i64 n = N;
    while (i <= n)
      if (n % i == 0)
        n /= i;
      else
        n %= i;
    if (n == 1)
      ans.insert(i);
  }
  cout << ans.size() << endl;
  return 0;
}
