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.Arrays;
import java.util.StringTokenizer;

public class Main {
	
	public static void main (String[ ]args) throws NumberFormatException, IOException{
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		final int  x= Integer.parseInt(br.readLine());
		StringTokenizer st = new StringTokenizer(br.readLine());
		int a [] = new int [x];
		int[] b = new int [x];
		int [] sorted = new int [x];
		
		
		for(int i=0;i<x;i++){
			b[i]=Integer.parseInt(st.nextToken());
			
		}
		int k =0 ;
		if (x%2==0){
			for(int i=0 ; i <a.length  ; i++){
				Arrays.sort(b);
				sorted[i]=b[k];
				sorted[++i]=b[x-1-k];
				k++;
				//System.out.println(b[x-1-k]);
			}
		}
		else{
			for(int i=0 ; i <a.length-1  ; i++){
				Arrays.sort(b);
				sorted[i]=b[k];
				sorted[++i]=b[x-1-k];
				k++;
				//System.out.println(b[x-1-k]);
			}
			int mid = x/2;
			
			sorted[x-1]= b[mid];
		}
//		for(int i=0 ; i <a.length  ; i++){
//			Arrays.sort(b);
//			sorted[i]=b[k];
//			sorted[++i]=b[x-1-k];
//			k++;
//			//System.out.println(b[x-1-k]);
//		}
		
		for(int i =0; i<x;i++){
			System.out.print(sorted[i]+" ");
			//System.out.print(b[i]);
		}
		
		
		
		
	}

}
