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 ff            first
#define ss            second
#define int           long long
#define pb            emplace_back
#define pii           pair<int,int>
#define vi            vector<int>
#define mii           map<int,int>
#define pq            priority_queue<int>
#define pqs           priority_queue<int,vi,greater<int>>
#define all(x)        x.begin() , x.end()
#define sortall(x)    sort(all(x));
#define clr(x)        memset(x , 0 , sizeof(x))
#define setbits(x)    __builtin_popcountll(x)
#define zrobits(x)    __builtin_ctzll(x)
#define sp(x,y)       fixed<<setprecision(y)<<x
#define query         int q; cin>>q; while(q--)
#define fo(a,n)       for(int i = a ; i < n ; i++)
#define forev(n,a)    for(int i = n ; i >= 0 ; i--)
#define deb(x)        cout << #x << " = " << x << endl
#define debar(x)      fo(i , x.size()) cout << x[i] <<" "
#define endl          '\n'
#define FIO           ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

const int MOD = 1e9 + 7;
const int INF = 1e18;
const int N   = 1e5 + 5;


void solve(){
    int n;
    cin >> n;
    vi ar(n+2);
    fo(0 , n + 2) cin >> ar[i];

    sort(all(ar));

    int mx = ar[n+1];
    int smx = ar[n];

    int s = 0;
    for(int i = 0 ; i < n ; i++) s += ar[i];

    if(s == mx || s == smx){
        for(int i = 0; i < n ; i++){
            cout << ar[i] << " ";
        }
        cout << endl;
        return;
    }

    s += smx;

    int i = 0;
    for(; i < n + 1 ; i++){
        if(s - ar[i] == mx){
            break;
        }
    }

    if(i == n + 1){
        cout << -1 << endl;
        return;
    }

    for(int j = 0 ; j < n + 1 ; j++){
        if(j == i) continue;
        else cout << ar[j] << " ";
    }
    cout << endl;
}

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