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++)
#define INF 1000000007
using namespace std;

int main(){
  long n;
  cin >> n;
  long ans = 1;
  long tmp = n;
  for(long i=2;i<=sqrt(n);i++){
    tmp = n;
    while(tmp%i==0)tmp /= i;
    if(tmp%i==1){
      ans++;
    }
  }
  for(long j=2;j<=sqrt(n);j++){
    if((n-1)%j==0 && (n-1)/j>sqrt(n))ans++;
  }
  cout << ans+(n>2) << endl;
  return 0;
}