Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

vector<int64_t> divisor(int64_t n){
  vector<int64_t> div;
  for(int64_t i=1;i*i<= n;++i) {
    if(n%i==0) {
      div.push_back(i);
      if(i*i!=n) div.push_back(n/i);
    }
  }
  sort(begin(div),end(div));
  return (div);
}


int main(){
  int64_t N;
  cin>>N;

  int64_t ans=0;
  vector<int64_t> v,w;
  v=divisor(N),w=divisor(N-1);
  for(int i=1;i<v.size();++i){
    int64_t n=N,m=v[i];
    while(n%m==0){
      n /= m;
    }
    n %= m;
    if(n==1) ++ans;
  }

  for(int i=1;i<w.size();++i){
    ++ans;
  }

  cout<<ans<<endl;
}