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>
#define ll long long int
#define ld long double
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define pii pair<int, int>
#define inf INT_MAX
#define endl '\n'
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define maxn 100000+5
#define eps 1e-10
#define ff first
#define ss second 
#define sz size
#define lnt length
#define here(x) cout << #x << " is " << x << endl;
#define setv(x, v) memset(x, v, sizeof(x))
#define prec(x) cout << fixed << setprecision(x);
#define fio ios_base::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);
using namespace std;
ll dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

void solve()
{
    int n;
    cin>>n;
    ll sum = 0;
    vector<ll>b(n + 2);
    for(int i=0;i<n + 2;i++)
        cin>>b[i], sum += b[i];
    sort(all(b));
    int pos1 = -1, pos2 = -1;
    for(int i=0;i<n + 2;i++)
    {
        if(i < n + 1 && sum - b[i] - b.back() == b.back())
        {
            pos1 = i;
            pos2 = b.sz() - 1;
            break;
        }
        else if(i == n + 1)
        {
            if(sum - b[i] - b[i - 1] == b[i - 1])
            {
                pos1 = i;
                pos2 = i - 1;
                break;
            }
        }
    }
    if(pos1 == -1)
        cout<<pos1<<endl;
    else
    {
        for(int i=0;i<n + 2;i++)
        {
            if(i != pos1 && i != pos2)
                cout<<b[i]<<" ";
        }
        cout<<endl;
    }
}

int main()
{
    fio;
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ll t;
    cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}