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<vector>
#include<algorithm>
using namespace std;
typedef long long int ll;
int main(){
	int N;
	vector<int> ans;
	ll sum = 0LL;
	int cnt = 0;
	cin >> N;
	for(int i = 2; i <= 30000; i++){
		if(i % 2 == 0 || i % 3 == 0){
			cnt++;
			ans.push_back(i);
			sum += i;
		}
		if(cnt == N) break;
	}
	if(sum % 2 != 0){
		for(int i = N - 1; i >= 0; i--){
			if(ans[i] % 3 == 0) ans[i] += 3;
			break;
		}
	}
	if(sum % 3 != 0){
		for(int i = N - 1; i >= 0; i--){
			if(ans[i] % 2 == 0){
				if(sum % 3 == 1){
					ans[i] += 2;
					break;
				}else{
					ans[i] += 4;
					break;
				}
			}
		}
	}
	for(int i = 0; i < N; i++){
		cout << ans[i];
		if(i == N - 1) cout << endl;
		else cout << " ";
	}
	return 0;
}