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 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> ii; 
typedef vector<ii> vii; 
typedef vector<int> vi;
typedef priority_queue<int, std::vector<int>, std::greater<int> >  rpq;
#define pb push_back
#define ll long long
#define F first
#define S second
#define eps 1e-9
#define M (ll)(1e9+7)
#define pi 3.14159265358979323846
#define ios ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);

int main()
{
    ios;
    ll t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        n+=2;
        ll a[n];
        for(int i=0;i<n;i++)
            cin>>a[i];
        sort(a, a+n);
        ll val=a[n-1];
        ll ans=-1, extra=-1;
        ll sum=0;
        for(int i=0;i<n-1;i++){
            sum+=a[i];
        }

        for(int i=0;i<n-1;i++){
            if(sum-a[i]==val){
                ans=val;
                extra=i;
            }
        }

        if(ans!=-1){
            for(int i=0;i<n-1;i++){
                if(i==extra)
                    continue;
                cout<<a[i]<<" ";
                
            }
            cout<<endl;
            continue;
        }

        extra=a[n-1];
        val=a[n-2];
        sum=0;
        for(int i=0;i<n-2;i++){
            sum+=a[i];
        }
        if(sum==val){
            ans=val;
        }

        if(ans==-1){
            cout<<ans<<endl;
        }else{
            for(int i=0;i<n-2;i++)
                cout<<a[i]<<" ";
            cout<<endl;
        }
    }
}