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)

#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <list>
#include <vector>
#include <queue>
#include <algorithm>
#include <climits>
#include <cstring>
#include <cmath>
#include <stack>
#include <iomanip>
#include <tuple>
#include <functional>
#include <cfloat>
#include <map>

#define int long long
#define CONTAINS(v,n) (find((v).begin(), (v).end(), (n)) != (v).end())
#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
#define ARY_SORT(a, size) sort((a), (a)+(size))
#define MAX(a,b) (((a) > (b)) ? (a) : (b))
#define MIN(a,b) (((a) < (b)) ? (a) : (b))
#define REMOVE(v,a) (v.erase(remove((v).begin(), (v).end(), (a)), (v).end()))
#define REVERSE(v) (reverse((v).begin(), (v).end()))
#define LOWER_BOUND(v,a) (lower_bound((v).begin(), (v).end(), (a)))
#define UPPER_BOUND(v,a) (upper_bound((v).begin(), (v).end(), (a)))
#define REP(i, n) for (int (i) = 0; (i) < (n); (i)++)
#define CONTAINS_MAP(m, a) (m).find((a)) != m.end()
using namespace std;

int N;
int ans[20000];

signed main()
{
	cin >> N;

	if (N == 3)
	{
		cout << "2 5 63" << endl;
		return 0;
	}

	if (N <= 15002)
	{
		int max = 0;
		int sum = 0;
		for (int i = 0; i < N - 3; i++)
		{
			ans[i] = (i + 1) * 2;
			max = MAX(ans[i], max);
			sum += ans[i];
		}
		for (int i = max + 2; i <= 30000; i += 2)
		{
			if(((sum + i) % 3) == 0)
			{
				ans[N - 3] = i;
				ans[N - 2] = 3;
				ans[N - 1] = 9;
				break;
			}
		}
	}
	else
	{
		int max = 0;
		int sum = 0;

		int m = 15000;
		if (N % 2 != 0)
		{
			m = 14999;
		}
		for (int i = 0; i < m; i++)
		{
			ans[i] = (i + 1) * 2;
			max = MAX(ans[i], max);
			sum += ans[i];
		}
		for (int i = 0; i < N - m; i++)
		{
			ans[i + m] = 3 * (i * 2 + 1);
		}
	}

	REP(i, N)
	{
		cout << (i ? " " : "") << ans[i];
	}
	cout << endl;
}