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;

#define ll long long
#define ld long double
#define pb push_back

void solve(){
    ll n; cin >> n;
    vector<ll> v(n + 2);
    ll sum = 0;
    for(ll i = 0; i < n + 2; i++){
        cin >> v[i];
        sum += v[i];
    }
    sort(v.begin(), v.end());
    ll a = -1, b = -1;
    for(ll i = 0; i < n + 2; i++){
        ll x = sum - 2 * v[i];
        ll l = 0, r = n + 2;
        while(r - l > 1){
            ll m = (r + l) / 2;
            if(v[m] > x){
                r = m;
            }
            else{
                l = m;
            }
        }
        if(l != i && v[l] == x){
            a = i;
            b = l;
            break;
        }
    }
    if(a == -1 && b == -1){
        cout << -1 << "\n";
    }
    else{
        for(ll i = 0; i < n + 2; i++){
            if(i != a && i != b){
                cout << v[i] << " ";
            }
        }
        cout << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cout.tie(0); cin.tie(0);
    //cout.precision(20); cout << fixed;
    ll t; cin >> t;
    while(t--){
        solve();
    }

    return 0;
}