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 <fstream>
using namespace std;
typedef long long int ll;
#define inf 1e18
#define mod 1000000007
#define mod1 998244353
#define mod2 999983
#define N 10000000
//***DON'T GET STUCK ON ONE APPROACH
ll power(ll a, ll n, ll m);
ll power(ll a, ll n, ll m)
{
    if (n == 0)
        return 1;
    ll x = power(a, n / 2, m);
    if (n % 2 != 0)
        return (((a * x) % m) * (x) % m) % m;
    else
        return (x * x) % m;
}
int32_t main()
{
    ll i, n, j, x, y, m, k, t, sum, count, cnt, b, c, d, j1, z, a, o, temp;
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> t;
    while (t--)
    {
        cin >> n;
        ll arr[n + 2];
        cnt = 0;
        map<ll, ll> p;
        for (i = 0; i < n + 2; i++)
        {
            cin >> arr[i];
            p[arr[i]]++;
            cnt += arr[i];
        }
        for (i = 0; i < n + 2; i++)
        {
            x = arr[i];
            if (p.find(cnt - x - x) != p.end())
            {
                if (cnt == x + x + x)
                {
                    if (p[x] == 1)
                    {
                        continue;
                    }

                    else
                    {
                        p[x] -= 2;
                        if (p[x] == 0)
                        {
                            p.erase(x);
                        }
                        break;
                    }
                }
                p[x]--;
                p[cnt - x - x]--;
                if (p[x] == 0)
                {
                    p.erase(x);
                }
                x = cnt - x - x;
                if (p[x] == 0)
                {
                    p.erase(x);
                }
                break;
            }
        }
        if (i == n + 2)
        {
            cout << "-1\n";
            continue;
        }
        for (auto i = p.begin(); i != p.end(); i++)
        {
            x = i->first;
            y = i->second;
            while (y--)
            {
                cout << x << " ";
            }
        }
        cout << "\n";
    }
    return 0;
}
