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>
using namespace std;

#define REP(i,n) for(int i=0;i<n;i++)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOR(i,m,n) for(int i=m;i<n;i++)
#define EPS (1e-7)

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<double,double> pdd;

int vx[]={0,1,0,-1};
int vy[]={1,0,-1,0};

int dx[]={1,0,-1,1,-1,1,0,-1};
int dy[]={1,1,1,0,0,-1,-1,-1};

const ll mod=1e9+7;
const int INF=INT_MAX/2;
const int MAX=1e9;

int gcd(int a,int b){
    return b!=0?gcd(b,a%b):a;
}

int main(){
	int n;
    cin>>n;
    int num=2;
    int cnt=0;
    int sum=0;
    while(cnt<n-2){
        if(num%2==0 || num%3==0){
            cout<<num<<" ";
            sum+=num;
            cnt++;
        }
        num++;
    }
    for(int i=num+1;i<=30000;i++){
        for(int j=i+1;j<=30000;j++){
            if((sum+i+j)%6==0){
                cout<<i<<" "<<j<<endl;
                return 0;
            }
        }
    }
	return 0;
}