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 CPP):

Solution:
#include <bits/stdc++.h>
#define tez ios_base::sync_with_stdio(false);cin.tie(NULL);
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mb make_pair
#define el "\n"
#define MAX 100005
using namespace std;
int main()
{
#ifndef ONLINE_JUDGE
	freopen("input1.txt", "r", stdin);
	freopen("output1.txt", "w", stdout);
#endif
	tez
	ll t;
	cin >> t;
	while (t--)
	{
		ll n;
		cin >> n;
		n += 2;
		vector <int> a(n);
		for (int i = 0; i < n; ++i)
			cin >> a[i];
		sort(a.begin(), a.end());
		ll sum = 0;
		for (int i = 0; i < n - 1; ++i)
			sum += a[i];
		int ind = -1;
		for (int i = 0; i < n - 1; ++i)
		{
			ll sumr = sum - a[i];
			if (sumr == a[n - 1] || sumr == a[i])
			{
				ind = i;
			}
		}
		if (ind != -1)
		{
			for (int i = 0; i < n - 1; ++i)
			{
				if (i != ind)
					cout << a[i] << " ";
			}
		}
		else
			cout << -1;
		cout << el;
	}
	return 0;
}