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>
using namespace std;
typedef long long ll;
typedef vector<ll> vint;
typedef vector<vint> vvint;
typedef vector<bool> vbool;
typedef vector<vbool> vvbool;
typedef pair<ll, ll> iPair;
#define ff first
#define ss second
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;
// Use 'cpp_int' instead of 'int'

int main(){
	ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

	// freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
	
	int t;
	cin >> t;
	while(t--){
		int n;
		cin >> n;
		n += 2;
		vint b(n);
		for(int i = 0; i < n; i++)
			cin >> b[i];
		sort(b.begin(), b.end());
		ll sum = accumulate(b.begin(), b.end(), 0ll);
		ll i1 = -1, i2 = -1;
		for(int i = 0; i < n; i++){
			//cout << sum - b[n - 1] - b[i] << " ";
			if(i != n - 1 && b[n - 1] == sum - b[n - 1] - b[i]){
				i1 = n - 1, i2 = i; 
				break;
			}
			if(i != n - 2 && b[n - 2] == sum - b[n - 2] - b[i]){
				i1 = n - 2, i2 = i; 
				break;
			}
		}
		if(i1 == -1){
			cout << "-1\n";
			continue;
		}
		for(int i = 0; i < n; i++){
			if(i != i1 && i != i2){
				cout << b[i] << " ";
			}
		}
		cout << "\n";
	} 
	
	return 0;
}