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)

#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma comment(linker, "/STACK:2000000")

#include "bits/stdc++.h"

using namespace std;

#define pb push_back
#define F first
#define S second
#define f(i,a,b)  for(int i = a; i < b; i++)
// #define endl '\n'
using ll = long long;
using db = long double;
using ii = pair<int, int>;

const int N = 2e5 + 5, LG = 19, MOD = 998244353;
const int SQ =320;
const long double EPS = 1e-7;
void print(vector<int> v){
  for(auto x : v)cout << x << ' ';
  exit(0);
}
int32_t main(){
#ifdef ONLINE_JUDGE
    ios_base::sync_with_stdio(0);
    cin.tie(0);
#endif

  int n;
  cin >> n;
  if(n == 3){
    cout << "2 5 63\n";
    return 0;
  }
  if(n <= 100){
    cout<<2<<' '<<3<<' '<<5<<' '<<20;
    f(i,0,n-4)cout<<' '<<30+i*30; return 0;
  }

  ll tot = 2 + 3 + 5;
  int cnt = n - 4;
  vector<int> v({2,3,5});
  for(int cur = 30000; cnt&&cur>=6;--cur){
    if((cur%2==0||cur%3==0||cur%5==0)&&(cur%30!=0)){
      v.push_back(cur);
      cnt--;
      tot += cur;
    }
  }


  while((tot%30)){
    tot-=v.back();
    v.pop_back();
  }
  for(int i = 0; v.size() < n; i++)v.pb((i+1)*30);

  print(v);

  return 0;
}