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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php

	$n = trim(fgets(STDIN));
	$i = 0;
	$c = 0;
	if($n == 3) echo "2 3 4";
	else{
    	while($c + 8 <= $n - ($n % 8)){
    	    for($j=0; $j<12; $j++){
    	       $i++;
        	    if($i%2==0 || $i%3==0){
        	        echo $i." ";
        	    } 
    	    }
    	    $c += 8;
    	}
    	if($c < $n){
    	    $r = $n - $c;
    	    if($r%2 == 1) echo ($i+6)." ";
    	    if($r > 1) echo ($i+2)." ".($i+4)." ";
    	    if($r > 3) echo ($i+3)." ".($i+9)." ";
    	    if($r > 5) echo ($i+8)." ".($i+10)." ";
    	}
	}
	
?>