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 (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
const ll MOD = 1000000007;
const ll INF = 10000000000000;
#define all(v) v.begin(), v.end()


int main(){
  ll N,M,sum=0;
  cin >> N;
  for(ll i=2;i*i<=N;i++){
    M=N;
    while(M%i==0){
      M /= i;
    }
    if(M%i==1){
      sum++;
    }
    if(N%i==1){
      sum++;
    }
    if(N==i*i+1){
      sum--;
    }
  }
  sum += 2;
  if(N==2){sum=1;}
  cout << sum << endl;
}

  