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 pb push_back
#define ff first
#define ss second
#define all(s) s.begin(), s.end()
#define FOR0(i, a) for (int i = 0; i < a; i++)
#define FOR(i, a, b) for (int i = a; i < b; i++)

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<bool> vb;

void solve() {
    int n;
    cin >> n;

    vll b(n + 2);
    FOR0(i, n + 2) cin >> b[i];

    sort(all(b));
    ll sum = 0;
    FOR0(i, n + 1) {
        sum += b[i];
    }
    int index = -1;
    FOR0(i, n + 1) {
        if (sum - b[i] == b[n + 1]) {
            index = i;
            break;
        }
    }

    if (index != -1) {
        FOR0(i, n + 1) {
            if (i == index) continue;
            cout << b[i] << " ";
        }
        cout << "\n";
        return;
    }
    if (sum - b[n] != b[n]) {
        cout << "-1\n";
        return;
    }
    
    FOR0(i, n) {
        cout << b[i] << " ";
    }
    cout << "\n";
}

int main (void) {
    cin.tie(0), cout.tie(0)->sync_with_stdio(0);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}