Problem: You are given a number n and an array b_1, b_2, …, b_{n+2}, obtained according to the following algorithm: 

  * some array a_1, a_2, …, a_n was guessed; 
  * array a was written to array b, i.e. b_i = a_i (1 ≤ i ≤ n); 
  * The (n+1)-th element of the array b is the sum of the numbers in the array a, i.e. b_{n+1} = a_1+a_2+…+a_n; 
  * The (n+2)-th element of the array b was written some number x (1 ≤ x ≤ 10^9), i.e. b_{n+2} = x; The 
  * array b was shuffled. 



For example, the array b=[2, 3, 7, 12 ,2] it could be obtained in the following ways: 

  * a=[2, 2, 3] and x=12; 
  * a=[3, 2, 7] and x=2. 



For the given array b, find any array a that could have been guessed initially.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second row of each test case contains n+2 integers b_1, b_2, …, b_{n+2} (1 ≤ b_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output: 

  * "-1", if the array b could not be obtained from any array a; 
  * n integers a_1, a_2, …, a_n, otherwise. 



If there are several arrays of a, you can output any.

Example

Input


4
3
2 3 7 12 2
4
9 1 7 1 6 5
5
18 2 2 3 2 9 2
3
2 6 9 2 1


Output


2 3 7 
-1
2 2 2 3 9 
1 2 6 

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;

public class CorruptedArray {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int test = scanner.nextInt();

        StringBuilder sb = new StringBuilder();
        for (int t = 0; t < test; t++) {
            int n = scanner.nextInt() + 2;

            Long[] arr = new Long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = scanner.nextLong();
            }

            Arrays.sort(arr);
            long sumAll = Arrays.stream(arr).mapToLong(e -> e).sum();

            long max1 = arr[n - 1];
            int max1Ind = n - 1;

            long max2 = arr[n - 2];
            int max2Ind = n - 2;

            // assume max1 is the sum.
            long rem1Sum = sumAll - max1;
            int result1 = -1;
            for (int i = 0; i < n; i++) {
                if (i == max1Ind) {
                    continue;
                }
                if (rem1Sum - arr[i] == max1) {
                    result1 = i;
                    break;
                }
            }

            long result2 = -1;
            // assume max2 is the sum.
            long rem2Sum = sumAll - max2;
            for (int i = 0; i < n; i++) {
                if (i == max2Ind) {
                    continue;
                }
                if (rem2Sum - arr[i] == max2) {
                    result2 = i;
                }
            }

            if (result1 == -1 && result2 == -1) {
                sb.append(-1).append(System.lineSeparator());
            }else{
                if (result1 >= 0) {
                    List<Long> resultArr = new ArrayList<>();
                    for (int i = 0; i < n; i++) {
                        if (i != max1Ind && i != result1) {
                            resultArr.add(arr[i]);
                        }
                    }
                    sb.append(resultArr.stream().map(e -> String.valueOf(e)).collect(Collectors.joining(" ")));
                    sb.append(System.lineSeparator());
                }else{
                    List<Long> resultArr = new ArrayList<>();
                    for (int i = 0; i < n; i++) {
                        if (i != max2Ind && i != result2) {
                            resultArr.add(arr[i]);
                        }
                    }
                    sb.append(resultArr.stream().map(e -> String.valueOf(e)).collect(Collectors.joining(" ")));
                    sb.append(System.lineSeparator());
                }
            }
        }
        System.out.println(sb);
    }
}