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
#define FAST ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define vll vector<ll>
#define pb push_back
using namespace std;
#define INF LLONG_MAX
const ll  N =1e9+7;
ll moduloexp(ll x, ll y , ll m) // o(logy)  returns x^y%m
{
    
    if(y==0)
    {
        return 1;
    }
    if(y%2==0)
    {
        return (moduloexp(x,y/2,m)*(moduloexp(x,y/2,m)))%m;
    }
    else
    {
        return((x%m)*(moduloexp(x,y-1,m)))%m;
    }
    
}
int main()

{ 
	FAST;
	     int tc;cin>>tc;
	 	while(tc--)
	 	{
	        ll n;
	        cin>>n;
	        ll a[n+2];
	        ll b[n+2];
	        for(ll i=0;i<n+2;i++)
	        {
	            cin>>a[i];
	            b[i]=a[i];
	        }
	        sort(a,a+n+2);
	        ll sum=0;
	        for(ll i=0;i<n;i++)
	        {
	            sum+=a[i];
	        }
	        
	        if(sum==a[n])
	        {
	            for(ll i=0;i<n;i++)
	            {
	                cout<<a[i]<<" ";
	            }
	            cout<<endl;
	        }
	        else if(sum==a[n+1])
	        {
	            for(ll i=0;i<n;i++)
	            {
	                cout<<a[i]<<" ";
	            }
	            cout<<endl;
	        }
	        else{
	            sum+=a[n];
	            ll j=0;
	            ll p=0;
	            for(ll i=0;i<n;i++)
	            {
	                if(sum-a[i]==a[n+1])
	                {
	                     j=i;
	                    p=1;
	                    break;
	                }
	            }
	            if(p)
	            {
	                for(ll i=0;i<n;i++)
	                {
	                    if(i!=j)
	                    {
	                        cout<<a[i]<<" ";
	                    }
	                }
	                cout<<a[n]<<endl;
	            }
	            else{
	                cout<<-1<<endl;
	            }
	        }
	 	}
}






