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 <list>
#include<stack>
#include<queue>
#include <vector>
#include <set>
#include<algorithm>
#include<math.h>
#include<stdlib.h>
#include<string>
#include <functional>
#include<fstream>

#define FOR(k,m,n) for(int (k)=(m);(k)<(n);(k)++)
#define REP(i,n) FOR((i),0,(n))
#define LL long long
#define CLR(a) memset((a),0,sizeof(a))
#define SZ(x) (int((x).size()))
#define WAITING(str) int str;std::cin>>str;
#define DEBUGING(str) cout<<str<<endl
using namespace std;

const LL MOD = 1000000007;// 10^9+7
const int INF = (1 << 30);


//変数
int N;
//ans2:3の倍数ではない、2の倍数の集合
//ans3:2の倍数ではない、3の倍数の集合
//ans5:2,3の倍数ではない、5の倍数の集合
vector<int> ans, ans2, ans3, ans5;
const int lim = 30000;




//サブ関数
//入力
void input()
{
	cin >> N;

}
void make_set()
{
	FOR(i, 1, lim / 2+1) {
		ans2.push_back(i * 2);
	}
	FOR(i, 1, lim / 3+1) {
		if (i % 2 == 0)continue;
		ans3.push_back(i * 3);
	}
}

void make_ans(int num2, int num3, int num5 = 0)
{
	REP(i, num2)ans.push_back(ans2[i]);
	REP(i, num3)ans.push_back(ans3[i]);
	REP(i, num5)ans.push_back(ans5[i]);
}


//計算
void calc()
{
	make_set();
	FOR(i, 1, 15001)FOR(j, 1, 5001) {
		if (i + j != N)continue;
		if (i % 3 == 1 || j % 2 != 0)continue;
		make_ans(i, j);
		//cerr << i << "  " << j << endl;
		return;
	}
	ans.push_back(2);
	ans.push_back(5);
	ans.push_back(63);
}


//出力
void output()
{
	for (auto num : ans) {
		cout << num << " ";
	}
	cout << endl;
}


//デバッグ
void debug()
{
	int N;
	cin>>N;
}


//メイン関数
int main()
{
	input();
	calc();
	output();
	debug();
	
	return 0;
}