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.util.*;
import java.io.*; 
public class B {
	public static void main(String[] args) throws IOException {
		in.init(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = in.nextInt();
		int[] a = new int[n];
		for(int i = 0; i < n; i ++)
			a[i] = in.nextInt();
		Arrays.sort(a);
		
		int[] ans = new int[n];
		int p2 = n-1;
		int offset = n%2;
		for(int i = 0; i < n; i ++)
		{
			if(p2<=i)
				break;
			if(i%2==1)
			{
				ans[i] = a[p2];
				ans[p2 - offset] = a[i];
			}
			else
			{
				ans[i] = a[i];
				ans[p2 - offset] = a[p2];
			}
			p2--;
		}
		
		if(n%2==1)
			ans[n-1] = a[n/2];
		for(int e: ans)
			out.print(e + " ");
		out.close();
	}

	public static class in {
		static BufferedReader reader;
		static StringTokenizer tokenizer;

		static void init(InputStream in) {
			reader = new BufferedReader(new InputStreamReader(in));
			tokenizer = new StringTokenizer("");
		}

		static String next() throws IOException {
			while (!tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(reader.readLine());
			}
			return tokenizer.nextToken();
		}

		static int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		static double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		static long nextLong() throws IOException {
			return Long.parseLong(next());
		}
	}
}
