Problem:
Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S = \{a_{1}, a_{2}, ..., a_{N}\} of distinct positive integers is called special if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

3 \leq N \leq 20000

Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

The elements must be distinct positive integers not exceeding 30000.

The gcd of all elements of S is 1, i.e. there does not exist an integer d > 1 that divides all elements of S.

S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

Sample Input 1

3

Sample Output 1

2 5 63

\{2, 5, 63\} is special because gcd(2, 5 + 63) = 2, gcd(5, 2 + 63) = 5, gcd(63, 2 + 5) = 7. Also, gcd(2, 5, 63) = 1. Thus, this set satisfies all the criteria.

Note that \{2, 4, 6\} is not a valid solution because gcd(2, 4, 6) = 2 > 1.

Sample Input 2

4

Sample Output 2

2 5 20 63

\{2, 5, 20, 63\} is special because gcd(2, 5 + 20 + 63) = 2, gcd(5, 2 + 20 + 63) = 5, gcd(20, 2 + 5 + 63) = 10, gcd(63, 2 + 5 + 20) = 9. Also, gcd(2, 5, 20, 63) = 1. Thus, this set satisfies all the criteria.

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

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cmath>

using namespace std;

int main(void){

  int N;
  cin >> N;

  if(N==3){
    cout << "2 5 63" << endl;
  }
  else if(N==4){
    cout << "2 3 4 9" << endl;
  }
  else if(N>4 && N<8){
    cout << "2 3 4 9 " ;
    for(int i=0;i<N-4;i++){
      cout << 6*(i+1) << " ";
    }
    cout << endl;
  }
  else{
    for(int i=0;i<N/8;i++){
      for(int j=0;j<3;j++){
	cout << 12*(i+1) - 6 - 3 - 1 + j  << " ";
      }
      cout << 12*(i+1) -6<< " ";
      for(int j=0;j<3;j++){
	cout << 12*(i+1) - 3 - 1 + j  << " ";
      }
      cout << 12*(i+1) << " ";      
    }
    int amari=N%8;
    //for(int i=0;i<amari;i++){
    //cout << 12*(N/8) + 6*(i+1) << " ";
    //}
    switch(amari){
    case 0:
      break;
    case 1:
      cout << 12*(N/8) + 6 ;
      break;
    case 2:
      cout << 12*(N/8) + 6 << " " << 12*(N/8+1);
      break;      
    case 3:
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +2 + j  << " ";
      }
      break;      
    case 4:
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +2 + j  << " ";
      }
      cout << 12*(N/8) + 6 ;
      break;      
    case 5:
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +2 + j  << " ";
      }
      cout << 12*(N/8) + 6 ;
      cout << 12*(N/8) + 12 ;
      break;      
    case 6:
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +2 + j  << " ";
      }
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +8 + j  << " ";
      }
      break;      
    case 7:
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +2 + j  << " ";
      }
      cout << 12*(N/8) + 6 ;
      for(int j=0;j<3;j++){
	cout << 12*(N/8) +8 + j  << " ";
      }
      break;      
    }
    
  }
  
  return 0;
}