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 ull long long int
#define pb push_back
#define vi vector<int>
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define MOD 1000000007
#define X first
#define Y second
#define sz(a) ((int)(a).size())
#define all(a) (a).begin(), (a).end()
#define pii pair<int,int>
using namespace std;
#define pr(x) cout << #x << " = " << x << endl;

void solve(int tc)
{
    int n;
    cin >> n;
    ll b[n+2];
    for(int i =0; i < n+2; i++)
    {
        cin >> b[i];
    }
    sort(b,b+n+2);
    ll sum = b[n+1];
    ll s=0;
    for(int i =0; i < n+1; i++)
    {
        s+= b[i];
    }
    int f=0;
    int idx =-1;
    // for(int i =0; i < n+2; i++)
    // {
    //     cout << b[i] << ", ";
    // }
    // pr(sum); pr(s); 
    for(int i =0; i < n+1; i++)
    {
        if(s-b[i]==sum)
        {
            f=1;
            idx=i;
            break;
        }
    }
    if(f==1)
    {
        for(int i =0; i < n+1; i++)
        {
            if(i!=idx) cout << b[i] << " ";
        }cout << endl; return;
    }

    sum = b[n];
    s= s- b[n];
    if(s==b[n])
    {
        for(int i =0; i < n; i++)
        {
            cout << b[i] << " ";
        }cout << endl; return;
    }
    cout << -1 <<endl; return;
}
int main()
{
    fast
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int tc=1;
    cin>>tc;

    for (int i = 1; i <= tc; i++)
    {
        solve(i);
    }

    return 0;
}