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)

#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <functional>
#include <utility>
#include <tuple>
#include <cctype>
#include <bitset>
#include <complex>
#include <cmath>
using namespace std;
//#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3fLL
#define MOD 1000000007
#define mp make_pair
#define mt make_tuple
#define pb push_back
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pint;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> tint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
int dx[8]={0,0,-1,1,1,1,-1,-1};
int dy[8]={-1,1,0,0,1,-1,1,-1};
const int SIZE=100;
//ここまでテンプレ
int main(){
	int N,count=0;
	cin>>N;
	vint ans;
	vint two,three,six;
	for(int i=1;i<=30000;i++){
		if(i%2==0 && i%3)
			two.pb(i);
		if(i%3==0 && i%2)
			three.pb(i);
		if(i%6==0)
			six.pb(i);
	}
	if(N>=15000){
		for(int i:two){
			ans.pb(i);
			count++;
		}
		for(int i:three){
			ans.pb(i);
			count++;
		}
		for(int i:six){
			if(count==N)
				break;
			ans.pb(i);
			count++;
		}
		for(int i=0;i<ans.size()-1;i++)
			cout<<ans[i]<<" ";
		cout<<ans.back()<<endl;
		//cout<<ans.size()<<endl;
		return 0;
	}
	if(N==3){
		cout<<"2 5 63"<<endl;
		return 0;
	}
	int TWO=0,THREE=0;
	ll STWO=0,STHREE=0;
	while(1){
		for(TWO;1;TWO++){
			ans.pb(two[TWO]);
			count++;
			STWO+=two[TWO];
			if(STWO%6==0)
				break;
		}
		int tend=THREE+2;
		for(THREE;THREE<tend;THREE++){
			ans.pb(three[THREE]);
			count++;
		}
		if(count>=N-5000){
			for(int i=0;i<=5000;i++){
				if(count==N)
					goto next;
				ans.pb(six[i]);
				count++;
			}
		}
	}
	next:
	for(int i=0;i<ans.size()-1;i++)
		cout<<ans[i]<<" ";
	cout<<ans.back()<<endl;

	ll sss=0;
	for(int i:ans)
		sss+=i;
	//cout<<sss<<" "<<ans.size()<<endl;
	return 0;
}