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
//package DivideAndConquer;

import java.io.*;
import java.util.*;

public class Main{
	
	static int n;
	static int[] arr;
	static boolean zSorted;
	static int[] perm;
	static boolean[] used;
	
	public static void main(String[] args) throws IOException, InterruptedException{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		n = Integer.parseInt(br.readLine());
		arr = new int[n];
		zSorted = false;
		perm = new int[n];
		used = new boolean[n];
		StringTokenizer st = new StringTokenizer(br.readLine());
		for(int i = 0; i<n; i++)
			arr[i] = Integer.parseInt(st.nextToken());
		solve(0);
		if(zSorted) {
			for(int i = 0; i<n; i++) {
				if(i==n-1)
					System.out.print(perm[i]);
				else
					System.out.print(perm[i]+" ");
			}
		}
		else System.out.println("Impossible");
	}
	
	public static void solve(int idx) {
		if(idx==n) {
			zSorted = true;
			return;
		}
		
		for(int i = 0; i<used.length; i++) {
			if(!used[i]) {
				if(idx>0) {
					if(idx%2!=0) {
						if(arr[i]>=perm[idx-1]) {
							used[i] = true;
							perm[idx] = arr[i];
							solve(idx+1);
							used[i] = false;
							if(zSorted) return;
						}
					}else {
						if(arr[i]<=perm[idx-1]) {
							used[i] = true;
							perm[idx] = arr[i];
							solve(idx+1);
							used[i] = false;
							if(zSorted) return;
						}
					}
				}else {
					used[i] = true;
					perm[idx] = arr[i];
					solve(idx+1);
					used[i] = false;
					if(zSorted) return;
				}
			}
		}
	}
	
}

 	      		 	 	 		 	  	  	 	 	 	