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 
#include<string>
#include<iostream>
#include<string>
#include<algorithm>
#include<math.h>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<iomanip> 
#include<unordered_map>
#include<bitset>
#define format(n) fixed<<setprecision(n)
#define ll long long
typedef unsigned long long ull;
#define  mod  1000000007 
const ll INF = 2e18;
const double EPS = 1e-9;
typedef long double ld;
#define ll_min LLONG_MIN
#define ll_max LLONG_MAX
#define endl "\n"
#define all(v) v.begin(),v.end()
#define sz(s)	(int)(s.size())
using namespace std;
void WEZaa() {
	std::ios_base::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
}
char arr[500][500];
int main()
{
	WEZaa();
	ll n, m, a, b, c = 0, x, y, z, t, sum = 0;
	cin >> t;
	while (t--)
	{
		cin >> n;
		n += 2;
		vector<int>arr(n+10);
		sum = 0;
		for (int i = 0; i < n; i++)
		{
			cin >> arr[i];
			sum += arr[i];
		}
		sort(arr.begin(), arr.begin() + n);
		bool bb = 0;
		sum -= arr[n - 1];
		for (int i = 0; i < n - 1; i++)
		{
			sum -= arr[i];
			if (sum == arr[n - 1])
			{
				bb = 1;
				for (int j = 0; j < n - 1; j++)
				{
					if (j != i)
						cout << arr[j] << " ";
				}
				cout << endl;
				break;
			}
			sum += arr[i];
		}
		if (bb == 0)
		{
			sum -= arr[n - 2];
			if (sum == arr[n - 2])
			{
				bb = 1;
				for (int j = 0; j < n - 2; j++)
				{
					cout << arr[j] << " ";
				}
				cout << endl;
			}
		}
		if (bb == 0)
			cout << -1 << endl;
	}
}
    