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: Accepted)

#include<iostream>
#include<algorithm>
#include<climits>
#include<vector>
#include<utility>
#include<cmath>
#include<string>
#include<cstring>
#include<map>
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define prin(arg) std::cout<<arg<<"\n"
#define prin2(arg1,arg2) std::cout<<arg1<<" "<<arg2<<"\n"
#define fill(arg,n) memset(arg,n,sizeof(arg))
using std::cin;
typedef long long ll;
typedef std::pair<int,int> pi;
typedef std::vector<int> vi;
const int INF=1e+9;
const ll INFLL=1e+17;
int N;
vi solve(){
  vi res;
  if(N==3){
    res.push_back(2);
    res.push_back(3);
    res.push_back(25);
    return res;
  }
  res.push_back(2);
  res.push_back(4);
  res.push_back(3);
  res.push_back(9);
  int j=1,s=4;
  while(s<N-1&&j<5000){
    res.push_back(6*j+2);
    res.push_back(6*j+4);
    j++;
    s+=2;
  }
  if(j!=5000){
    if(N%2==1){
      res.push_back(6);
      return res;
    }
    else{
      return res;
    }
  }
  j=1;
  while(s<N-1&&j<2500){
    res.push_back(12*j+3);
    res.push_back(12*j+9);
    j++;
    s+=2;
  }
  if(j!=2500){
    if(N%2==1){
      res.push_back(6);
      return res;
    }
    else{
      return res;
    }
  }
  j=1;
  while(s<N){
    res.push_back(6*j);
    j++;
    s++;
  }
  return res;
}
int main(){
  cin>>N;
  vi res=solve();
  int rs=res.size();
  rep(i,rs) std::cout<<res[i]<<" ";
  std::cout<<"\n";
  return 0;
}