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 endl '\n'
#define fi first
#define se second
#define MOD(n,k) ( ( ((n) % (k)) + (k) ) % (k))
#define FOR(i,n) for (int i = 0; i < n; i++)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<ii> vii;
typedef vector<vii> vvii;

int n, a, b;
bool f = 0;
ll suma = 0;

int main () {
    ios_base::sync_with_stdio(0); cin.tie(0);
    
 	cin >> n;
 	a = 15000;

 	if (n == 19999) {
 		f = 1;
 		n -= 12;
 	}

 	if (n == 3) {
 		cout << "2 5 63" << endl;
 		return 0;
 	}
 	if (n == 4) {
 		cout << "2 5 20 63" << endl;
 		return 0;
 	}
 	if (n == 6) {
 		cout << "2 5 20 63 30 60" << endl;
 		return 0;
 	}

 	while (a) {
 		if (a < n) {
 			b = n - a;

 			if (b % 2 == 0 && b * 6 - 3 <= 30000)
 				break;
 		}

 		a -= 3;
 	}

 	assert(a > 0 && b > 0 && b * 6 - 3 <= 30000);

 	int x = 2, y = 3;
 	while (a--) {
 		assert(x <= 30000);
 		cout << x << " ";
 		suma += x;
 		x += 2;
 	}

 	while (b--) {
 		assert(y <= 30000);
 		cout << y << " ";
 		suma += y;
 		y += 6;
 	}

 	if (f) {
	 	a = 12, x = 17;
	 	while (a--) {
	 		assert(x <= 30000 && x % 2 && x % 3);
	 		cout << x << " ";
	 		suma += x;
	 		x += 17 * 6;
	 	}
 	}

 	cout << endl;
 	//cout << suma << endl;

    return 0;
}