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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int64_t N;
  cin >> N;
  int res = 0;
  if (N == 2) {
    cout << 1 << '\n';
    return 0;
  }
  for (int64_t i = 2; i * i < N + 1; i++) {
    if ((N - 1) % i == 0) {
      res += (i * i == N - 1 ? 1 : 2);
    }
    if (N % i == 0) {
      int64_t x = N;
      while (x % i == 0) {
        x /= i;
      }
      if (x % i == 1) res++;
    }
  }
  cout << res + 2 << '\n';
  return 0;
}
