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>
#include <unordered_map>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef pair<int, ii> iii;
typedef vector<ii> vii;
typedef set<int> si;
typedef map<string, int> msi;
typedef map<string, string> mss;
#define re return
#define clr(x, v) memset(x, v, sizeof(x))
#define mp make_pair
#define pb push_back
#define mt make_tuple
#define eb emplace_back
#define bn begin()
#define nd end()
#define X first
#define Y second
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define UNIQUE(c) (c).resize(unique(ALL(c)) - (c).begin())
#define NumofDigits(n) ((int)log10(n) + 1)
#define sc1(x) scanf("%d", &x)
#define sc2(x, y) scanf("%d%d", &x, &y)
#define sc3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define sz size()
#define rep(i, v) for (int i = 0; i < (int)v.sz; i++)
#define lp(i, a, b) for (int i = int(a); i < int(b); i++)
#define lpi(i, b, a) for (int i = int(b); i >= a; i--)
#define watch(x) cout << (#x) << " is " << (x) << endl
#define print(ans) cout << ans << endl
#define print2(a, b) cout << a << ' ' << b << endl
#define EPS 1e-10
#define MAX 1000005
#define INF 1e18
ll MOD = 1e9 + 7;
const double PI = acos(-1.0);
#define print_prec(ans, num) cout << fixed << setprecision(num) << (ans) << endl
#define IO                            \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL)
const int dx[] = {1, -1, 0, 0, 0, 0};
const int dy[] = {0, 0, 1, -1, 0, 0};
const int dz[] = {0, 0, 0, 0, 1, -1};
const int N = 2e5 + 5;

int main()
{
    //freopen("read.txt", "r", stdin);
    //freopen("write.txt", "w", stdout);
    IO;
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        n += 2;
        vll a(n);
        ll sm = 0;
        lp(i, 0, n) cin >> a[i], sm += a[i];
        sort(all(a));
        int x = -1, e = -1;
        lp(i, 0, n)
        {
            if (i != n - 1)
            {
                if (sm - a[i] - a[n - 1] == a[n - 1])
                {
                    x = i;
                    e = n - 1;
                    break;
                }
            }
            else
            {
                if (sm - a[i] - a[n - 2] == a[n - 2])
                {
                    x = i;
                    e = n - 2;
                    break;
                }
            }
        }
        if (x == -1)
        {
            cout << -1 << endl;
        }
        else
        {
            lp(i, 0, n) if (i != x && i != e) cout << a[i] << " ";
            cout << endl;
        }
    }
}