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;

/**
 * Created by n on 09.05.2016.
 */
public class B652 {

    public static void main(String[] args) {

        MyScanner in = new MyScanner();
        int n  = in.nextInt();
        int[] arr = new int[n];
        for(int i = 0; i < n; i++){
            arr[i] = in.nextInt();
        }

        Arrays.sort(arr);

        int[] ans = new int[n];
        int k = n /2;

        int p = 0;
        int q = n - 1;
        for(int i = 0; i < n;i++){
            if (i % 2 == 0){
                ans[i] = arr[p];
                p++;
            }
            else{
                ans[i] = arr[q];
                q--;
            }
        }

        for(int i = 0; i < n; i++){
            System.out.print(ans[i] + " ");
        }
        System.out.println();



    }

    private static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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