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 <iostream>
#include <string>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <numeric>
#include <cmath>
#include <cassert>

using namespace std;

typedef long long int ll;
typedef pair<int,int> P;

#define rep(i,n) for(int i=0;i<(n);++i)
#define all(x) x.begin(),x.end()

const ll mod = 1e9+7;
const ll INF = 1e9;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int main()
{
	int n;
	cin>>n;
	set<int>ans;
	if(n==3){
		ans = {2,5,63};
	}else if(n==4){
		ans={2,5,20,63};
	}else if(n==5){
		ans={2,5,20,30,63};
	}else{
		ans={2,3,4};
		n -= 3;
		for (int x = 6; x <= 30000; x++){
			if(x%6==1||x%6==5) continue;
			ans.insert(x);
			n--;
			if(n==0)break;
		}
		int sum = accumulate(all(ans),0);
		if(sum%6==2){
			ans.erase(8);
			ans.insert(30000);
		}else if(sum%6==3){
			ans.erase(9);
			ans.insert(30000);
		}else if(sum%6==4){
			ans.erase(9);
			ans.insert(30000-2);
		}
	}

	for(auto x:ans)cout<<x<<" ";
	cout<<endl;
	return 0;
}