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 
/*    Author : Prakhar Rai    */
#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define LB(x,num) lower_bound(x.begin(),x.end(),num) - x.begin()
#define UB(x,num) upper_bound(x.begin(),x.end(),num) - x.begin()
#define BS(x,num) binary_search(x.begin(),x.end(),num)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define vci vector<int>
#define vcll vector<ll>
#define vcd vector<long double>
#define line(x) sort(x.begin(),x.end())
#define all(x) x.begin(),x.end()
#define newl "\n"
#define vc vector
#define loop(i,a,b) for(int i = a; i < b; i++)
#define lol(i,a,b) for(ll i = a; i < b; i++)
#define lod(i,a,b) for(ld i = 0; i < b; i++)
#define mod 1000000007
#define read(v,n) lol(i,0,n) {ll x; cin >> x; v.pb(x);}
#define run(a,x) for(auto x : a)
#define yes "YES"
#define no "NO"
ll cnt;
using namespace std;
// ll dp[10000000];
// memset(dp, -1, sizeof(dp));

void solve() {
	ll n; cin >> n;
	vcll v;
	read(v, n + 2);
	sort(all(v));
	ll sum = 0;

	lol(i, 0, n) {
		sum += v[i];
	}

	if (sum == v[n]) {
		lol(i, 0, n) {
			cout << v[i] << ' ';
		}
		cout << newl;
		return;
	}

	sum += v[n];

	lol(i, 0, n + 1) {
		if (sum - v[i] == v[n + 1]) {
			lol(j, 0, n + 1) {
				if (i == j) {
					continue;
				} else {
					cout << v[j] << ' ';
				}
			}
			cout << newl;
			return;
		}
	}

	cout << -1 << newl;
	return;
}

// FILE WITH TEST CASE

void init() {
#ifndef ONLINE_JUDGGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
}

int main() {
// 	init();
	ios_base::sync_with_stdio(false); cin.tie(0);
	int t; cin >> t;
	while (t--)
		solve();
}
/* This is my journey and i shall endure the work to reach the top .*/