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 ll = long long;

int main(){
  ll N;cin >> N;
  if(N==2){
    cout << 1 << endl;
    return 0;
  }else if(N==3){
    cout << 2 << endl;
    return 0;
  }
  ll ans = 2;
  for(ll i=2;i*i<=N;i++){
    if(N%i==0){
      ll x = N;
      while(x%i==0){
        x /= i;
      }
      if(x%i==1){
        //cout << "a " << i <<  endl;
        ans ++;
      }
    }
    if((N-1)%i==0){
      if(i*i==N-1){
        ans ++;
      }else{
        ans += 2;
      }
    }
  }
  cout << ans << endl;
}