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 "bits/stdc++.h"
using namespace std;

#define int long long
#define pb push_back
#define f(i,a,n) for(int i=a ; i<n ; i++)
#define rf(i,n,a) for(int i=n ; i>=a ; i--)
#define F first
#define S second
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int inf = 1e9;
const int inf64 = 1e18;
const int MOD = inf + 7;

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

    int a[n+2];
    f(i,0,n+2) cin >> a[i];

    sort(a, a+n+2);
    int sum = 0;

    f(i,0,n+2) sum += a[i];

    f(i,0,n+1){
        // cout << a[i] << " " << sum << " " << sum-a[i] << " " << a[n+1] << "\n";
        if(sum - a[i] - a[n+1] == a[n+1]){
            f(j,0,i) cout << a[j] << " ";
            f(j,i+1,n+1) cout << a[j] << " ";
            cout << "\n";
            return ;
        }
    }

    if(sum - a[n+1] == 2 * a[n]){
        f(i,0,n) cout << a[i] << " ";
        cout << "\n";
        return ;
    }
    cout << "-1\n";
}

int32_t main() {
    fast;
    int t = 1;
    cin >> t;
    while (t--)  solve();
}