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.io.PrintWriter;
import java.util.StringTokenizer;

public class P4_ZSort {
	
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		StringTokenizer st;
		int n = Integer.parseInt(br.readLine());
		int[] a = new int[n];
		st = new StringTokenizer(br.readLine());
		for(int i = 0 ; i < n ; i++)
			a[i] = Integer.parseInt(st.nextToken());
		devide(a,0,n-1);
		for(int i = 0 ; i < n>>1 ; i++){
			pw.print(a[i] + " " + a[n-i-1]);
			if(i < (n>>1)-1)
				pw.print(" ");
		}
		if(n%2 == 1)
			pw.print(" " + a[n>>1]);
		pw.println();
		pw.flush();
	}
	
	private static void devide(int[] a, int s, int e) {
		if(s == e)
			return;
		int mid = (e+s)>>1;
		devide(a , s , mid);
		devide(a , mid+1 , e);
		merge(a , s , e);
	}

	private static void merge(int[] a, int s, int e) {
		int mid = (e+s)>>1 , i = s , j = mid+1 , length = e - s + 1;
		int[] x = new int[length];
		for(int l = 0 ; l < length ; l++){
			if(i > mid || (j <= e && a[i] > a[j]))
				x[l] = a[j++];
			else
				x[l] = a[i++];
		}
		for(int l = 0 ; l < length ; l++)
			a[l+s] = x[l];
	}

}
