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 ll long long
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define deb(x) cout<<#x<<"="<<x<<endl;
#define endl '\n'
#define M 1000000007
#define int long long
#define F first
#define S second
#define INF 1e18
#define N 1000005

using namespace std;


ll b[200005];

void solve()
{
	ll n;
	cin >> n;

	ll actual = 0;
	for (int i = 0; i < n + 2; ++i)
	{
		cin >> b[i];
		actual += b[i];
	}
	sort(b, b + n + 2);
	ll sum = b[n + 1];
	ll act = actual;
	act -= sum;
	for (int i = 0; i < n + 1; ++i)
	{
		act -= b[i];
		if (act == sum)
		{
			for (int j = 0; j < n + 2; ++j)
			{
				if (j == n + 1)continue;
				if (j == i)continue;
				cout << b[j] << " ";
			}
			cout << endl;
			return;
		}
		act += b[i];
	}

	sum = b[n];

	act = actual;
	act -= sum;
	for (int i = 0; i < n + 2; ++i)
	{
		if (i == n)continue;
		act -= b[i];
		if (act == sum)
		{
			for (int j = 0; j < n + 2; ++j)
			{
				if (j == n)continue;
				if (j == i)continue;
				cout << b[j] << " ";
			}
			cout << endl;
			return;
		}
		act += b[i];
	}

	cout << -1 << endl;

}

int32_t main()
{
	IOS
	ll T = 1;
	cin >> T;
	for (ll i = 1; i <= T; ++i)
	{
		// cout<<"Case #"<<i<<": ";
		solve();
	}
	return 0;
}