A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:

  1. ai ≥ ai - 1 for all even i, 
  2. ai ≤ ai - 1 for all odd i > 1. 



For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.

Can you make the array z-sorted?

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".

Examples

Input

4
1 2 2 1


Output

1 2 1 2


Input

5
1 3 2 2 5


Output

1 5 2 3 2
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;

import javax.swing.text.StyledEditorKit.ForegroundAction;


public class ZSort {

	/**
	 * @param args
	 * @throws IOException 
	 * @throws NumberFormatException 
	 */
	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub
		BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		int length=Integer.parseInt(bf.readLine());
		String [] input=bf.readLine().split(" ");
		ArrayList<Integer> my_array=new ArrayList<Integer>();
		for (int i = 0; i < input.length; i++) {
			my_array.add(Integer.parseInt(input[i]));
			}
		
		Collections.sort(my_array);
		int [] result=new int[length];
		int midium=(int) Math.ceil(length/2.0);int last=length-1;
		int position=0;
		for (int i = 0; i < midium; i++) {
			result[position]=my_array.get(i);
			position+=2;
				}
		position=1;int cond;
		if(length%2==0){
			cond=midium;
		}
		else{
			cond=midium-1;
		}
		for (int i = 0; i < cond; i++) {
			result[position]=my_array.get(last);
			--last;
			position+=2;
		}
		for (int i = 0; i < result.length; i++) {
			System.out.print(result[i]+" ");
		}
		}
	
	
	
		

	}


