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)

#include <bits/stdc++.h>
#define ll long long
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	if (N == 3)
	{
		cout << "2 5 63" << endl;
		return 0;
	}
	else if (N == 4)
	{
		cout << "2 5 20 63" << endl;
		return 0;
	}
	
	int a, b;
	// aは3の倍数を出力する回数
	// bは(2の倍数) && !(3の倍数)を出力する回数
	if (N % 2 == 1)
	{
		a = min((N / 4 - 1) * 4 + 3, 9999);
		b = N - a;
	}
	else
	{
		a = min((N - 1) / 4 * 4, 100000);
		b = N - a;
	}
	for (int i = 1; i <= a; i++)
		cout << 3 * i << " ";
		
	int cnt = 0, x = 2;
	while (true)
	{
		if (x % 3 == 0)
		{
			x += 2;
			continue;
		}
		else
		{
			cout << x << (cnt != b - 1 ? " " : "");
			if (cnt == b - 1)
				break;
			else
				cnt++;
			x += 2;
		}
	}
}