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<bits/stdc++.h>
using namespace std;
#define int long long 
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define F first
#define S second
int const M=2e5+10,inf=1e9+10,mod=1e9+7;
int a[M],n;
int ans[M];
bool mark[M];
int pnt=1;
main()
{
	int now=0;
	cin>>n;
	int tmp=0;
	if(n==3)return cout<<"2 5 63",0;
	for(int i=1;true;i++)
	{	
		if(i%2==0 || i%3==0)
		{
			ans[pnt++]=i;
			mark[i]=1;
			tmp+=i;
		}
		if(pnt==n-1)break;
	}		
	if(tmp%3!=0)
	{
		int hlp=3-tmp%3;
		for(int j=2;true;j+=2)
		{
			if(!mark[j] && j%3==hlp)
			{
				ans[pnt++]=j;
				mark[j]=1;
				break;
			}
		}
	}
	if(tmp%2==1)
	{
		for(int j=3;true;j+=6)
		{
			if(!mark[j]){
				ans[pnt++]=j;
				mark[j]=1;
				break;
			}
		}
	}
	for(int j=6;true;j+=6)
	{
		if(pnt==n+1)break;
		if(!mark[j])
		{
			ans[pnt++]=j;
		}
	}
	for(int i=1;i<pnt;i++)
	{
		cout<<ans[i]<<" ";
	}
}