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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Z_Sort {

	
	public static void main(String[] args)throws IOException {
		
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a [] = new int[n];
		for (int i = 0; i < a.length; i++) 
			a[i] = sc.nextInt();
		Arrays.sort(a);
		if(n > 2)
			for (int i = 2; i < a.length; i+=2) {
				int tmp = a[i-1];
				a[i-1] = a[i];
				a[i] = tmp;
			}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < a.length; i++){
			sb.append(a[i]);
			sb.append(" ");
		}
		System.out.println(sb);
		
			
		
		
	}
	
	
	
	static class Scanner 
	{
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {

			br = new BufferedReader(new InputStreamReader(s));

		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public boolean ready() throws IOException {
			return br.ready();
		}
	}
}
