#include<bits/stdc++.h>
using namespace std;

#define ll long long 
ll ispresent(ll k,vector<ll>arr)
{
    for(ll i = 0;i<arr.size()-1;i++)
    {
        if(k == arr[i]) return i;
    }
    
    return -1;
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif    
      
    ll t;
    cin>>t;

    while(t--)
    {
        ll n;
        cin>>n;
        
        vector<ll>arr(n+2);

        for(ll i = 0;i<n+2;i++)
        {
            cin>>arr[i];
        }

        sort(arr.begin(),arr.end());

        ll sum = 0;

        for(ll i = 0;i<n;i++)
        {
            sum += arr[i];
        }
        if(arr[n] == sum)
        {
            for(ll i = 0;i<n;i++)
            {
                cout<<arr[i]<<" ";
            }
        }
        else
        {
            sum += arr[n];
            ll k = sum - arr[n+1];
            ll p = ispresent(k,arr);

            if(p!=-1)
            { 
            for(ll i = 0;i<n+1;i++)
            {
                if(i!=p)
                cout<<arr[i]<<" ";
            }
            }
            else
            cout<<"-1";
        }
        cout<<endl;

        


        
    }
}
### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

#define ll long long 
ll ispresent(ll k,vector<ll>arr)
{
    for(ll i = 0;i<arr.size()-1;i++)
    {
        if(k == arr[i]) return i;
    }
    
    return -1;
}
int main()
{
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif    
      
    ll t;
    cin>>t;

    while(t--)
    {
        ll n;
        cin>>n;
        
        vector<ll>arr(n+2);

        for(ll i = 0;i<n+2;i++)
        {
            cin>>arr[i];
        }

        sort(arr.begin(),arr.end());

        ll sum = 0;

        for(ll i = 0;i<n;i++)
        {
            sum += arr[i];
        }
        if(arr[n] == sum)
        {
            for(ll i = 0;i<n;i++)
            {
                cout<<arr[i]<<" ";
            }
        }
        else
        {
            sum += arr[n];
            ll k = sum - arr[n+1];
            ll p = ispresent(k,arr);

            if(p!=-1)
            { 
            for(ll i = 0;i<n+1;i++)
            {
                if(i!=p)
                cout<<arr[i]<<" ";
            }
            }
            else
            cout<<"-1";
        }
        cout<<endl;

        


        
    }
}
```