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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  ll n;
  cin >> n;
  int ans = 1;
  for (ll k = 2; k * k <= n; k++) {
    ll n_ = n;
    if (n_ % k) continue; 
    while (n_ % k == 0) n_ /= k;
    if (n_ % k == 1) ans++;
  }
  n--;
  if (n > 1) ans++;
  for (ll k = 2; k * k <= n; k++) {
    if (n % k == 0) ans += 2;
    if (k * k == n) ans--;
  }

  cout << ans << endl;
  return 0;
}