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>
using namespace std;
typedef long long ll;
typedef pair<int, int> pi;
const int MAX = 2e9;

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
        map<ll, int>mp;
        int n;
        cin >> n;
        n += 2;
        vector<ll>v(n);
        vector<bool>bv(n);
        ll sum = 0;
        ll myx = -2e16, mysum = -2e16;
        for (ll& i : v) {
            cin >> i;
            mp[i]++;
            sum += i;
        }
        for (int i = 0; i < n; i++) {
            sum -= v[i];
            if (--mp[v[i]] == 0) {
                mp.erase(v[i]);
            }
            if (prev(mp.end())->first * 2 == sum) {
                myx = i;
                mysum = prev(mp.end())->first;
                bv[i] = 1;
                break;
            }
            mp[v[i]]++;
            sum += v[i];
        }
        if (myx == -2e16) {
            cout << -1 << "\n";
        }
        else {
            for (int i = 0; i < n; i++) {
                if (v[i] == mysum && i != myx) {
                    bv[i] = 1;
                    break;
                }
            }
            for (int i = 0; i < n; i++) {
                if (!bv[i])cout << v[i] << " ";
            }
            cout << "\n";
        }
    }
}