You're given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn't equal to the sum of the last n elements?

Input

The first line contains an integer n (1 ≤ n ≤ 1000), where 2n is the number of elements in the array a.

The second line contains 2n space-separated integers a_1, a_2, …, a_{2n} (1 ≤ a_i ≤ 10^6) — the elements of the array a.

Output

If there's no solution, print "-1" (without quotes). Otherwise, print a single line containing 2n space-separated integers. They must form a reordering of a. You are allowed to not change the order.

Examples

Input


3
1 2 2 1 3 1


Output


2 1 3 1 1 2

Input


1
1 1


Output


-1

Note

In the first example, the first n elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4. The sums aren't equal.

In the second example, there's no solution.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Submit {

    public static void main(String[] args) throws IOException {
        InputStreamReader inputStreamReader = new InputStreamReader(System.in);
        BufferedReader bf = new BufferedReader(inputStreamReader);


        int n = Integer.valueOf(bf.readLine());
        Integer[] input = new Integer[2 * n];
        String[] elm = bf.readLine().split(" ");
        StringBuilder res = new StringBuilder();

        for (int i = 0; i < 2 * n; i++) {
            input[i] = (Integer.valueOf(elm[i]));
        }

        Arrays.sort(input, Collections.reverseOrder());

        int sum1 = 0, sum2 = 0;

        for (int i = 0; i < 2 * n; i++) {
            if (i < n)
                sum1 += input[i];
            if (i >= n)
                sum2 += input[i];
        }

        if (sum1 == sum2)
            System.out.println(-1);
        else {
            for (int i = 0; i < 2 * n; i++) {
                res.append(input[i]);
                res.append(" ");
            }
            res.deleteCharAt(res.length() - 1);
            System.out.println(res);
        }
    }
}
