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 <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <map>
using namespace std;
const int N = 2e5+10;//2e5表示2*10的5次方
typedef long long ll;
ll b[N];
map<ll,ll> m;
map<ll,ll>::iterator iter;
int main()
{
    std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int t;
    cin >> t;
    while(t--){
        m.clear();
        int n;
        cin >> n;
        for(int i=1;i<=n+2;i++){
            cin >> b[i];
            m[b[i]]++;
        }
        //1~n+2
        sort(b+1,b+n+3);//排序，小相加等于大
        m[b[n+2]]--;
        ll sum = 0;
        for(int i=1;i<=n;i++){
            sum+=b[i];//前n项和
        }
        if(sum==b[n+1]){
            for(int i=1;i<=n;i++) cout << b[i] << ' ';
            cout << '\n';
        }else{
            sum+=b[n+1];
            ll tmp = sum - b[n+2];
            if(m[tmp]>=1){
                m[tmp]--;
                for(iter=m.begin();iter!=m.end();iter++){
                    while(iter->second){//存在时，有几个重复的数就都输完
                        m[iter->first]--;
                        cout << iter->first << ' ';
                    }
                }
                cout << '\n';
            }else{
                cout << -1 << '\n';
            }
        }
    }
    return 0;
}

			 	     	   							  		  		