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 

// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>
using namespace std;

# define fast ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define endl "\n"
#define all(v) v.begin(),v.end()
#define mod 1000000007
#define len(s) int(s.size())
#define na "NO"
#define ha "YES"
#define pb push_back
#define s second
#define mem(a,val) memset(a,val,sizeof(a))
#define f first
#define rev(a) reverse(all(a))
#define ep emplace_back
#define read(arr, n)  for (int i = 0; i < n; i++)  cin >> arr[i];
#define read2(arr, n , m)  for (int i = 0; i < n; i++) for(int j=0; j<m; j++) cin >> arr[i][j];
typedef long long ll;
#define vi vector<int>




int d4[5] = { -1, 0, 1, 0, -1};

void solve()
{
	int n; cin >> n;
	n += 2;
	vector<ll> a(n);
	read(a, n);
	sort(all(a));

// x >= sigma ai
	ll sm = 0 ;
	for (int i = 0; i < n - 2; i++) sm += a[i];

	if (sm == a[n - 2]) {
		for (int i = 0; i < n - 2; i++) cout << a[i] << " ";
		cout << endl;
		return;
	}
// x < sigma ai
	sm = 0 ;
	for (int i = 0; i < n - 1; i++ ) sm += a[i];
	for (int i = 0; i < n - 1; i++) {
		if (sm - a[i] == a[n - 1]) {
			for (int j = 0 ; j < n - 1; j++) if (i != j) cout << a[j] << " ";

			cout << endl;
			return;
		}
	}
	cout << -1 << endl;

}


int main()
{
	fast;

	int t; cin >> t; while (t--)
		solve();

#ifndef ONLINE_JUDGE
	cerr  << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
#endif
}