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 <cstdio>
using namespace std;
const int maxn=1e5+10;

int n,s;
int a[maxn];

int nxt(int p)
{
	if (p%6==0) return p+2;
	else if (p%6==2) return p+1;
	else if (p%6==3) return p+1;
	else return p+2;
}

int main()
{
	scanf("%d",&n);
	if (n<=5)
	{
		if (n==3) puts("2 5 63");
		else if (n==4) puts("2 5 20 63");
		else puts("2 5 20 30 63");
		return 0;
	}
	
	int now=0;
	s=2;
	for(int i=1;i<=n;i++)
	{
		now=nxt(now);
		a[i]=now;
		s=(s+now)%6;
	}
	
	if (s==2)
	{
		do
		{
			now=nxt(now);
		} while(now%6!=0);
		a[5]=now;
	}
	else if (s==3)
	{
		do
		{
			now=nxt(now);
		} while(now%6!=0);
		a[6]=now;
	}
	else 
	{
		do
		{
			now=nxt(now);
		} while(now%6!=4);
		a[6]=now;
	}
	
	for(int i=1;i<=n;i++) printf("%d ",a[i]);
	puts("");
	return 0;
}