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)

/*input
3
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define REP(i,j,k)     for(int i = j ; i < k ; ++i)
#define RREP(i,j,k)    for(int i = j ; i >=k ; --i)
#define A    first
#define B    second
#define mp   make_pair
#define pb   emplace_back
#define PII pair<int , int>
#define MEM(i,j)   memset(i , j , sizeof i)
#define ALL(i)     i.begin() , i.end()
#define DBGG(i,j)     cout << i << " " << j << endl
#define DB4(i,j,k,l)  cout << i << " " << j << " " << k << " " << l << endl
#define IOS cin.tie(0) , cout.sync_with_stdio(0)
#define endl "\n"
///------------------------------------------------------------
#define MAX 30000
#define INF 0x3f3f3f3f


int N = 60 , n;
int32_t main(){
	IOS;
	cin >> n;
	vector<int> sol;
	int sum = 0;
	REP(i , 1 , MAX + 1){
		if(sol.size() == n - 1) break;
		else if(__gcd(i , N) != 1) sol.pb(i) , sum += i;
	}
	REP(i , sol.back() + 1 , MAX + 1){
		if((sum + i) % 2 == 0 && (sum + i) % 3 == 0 && (sum + i) % 5 == 0){
			sol.pb(i);
			break;
		}
	}
	while(sol.size() < n);
	for(auto to : sol) cout << to << " "; cout << endl;
    return 0;
}