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>
using namespace std;
#define fi(i, l, r) for (int i = l; i < r; i++)
typedef long long ll;
ll a[200010];
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        scanf("%d", &n);
        n+=2;
        ll sum=0;
        map<ll,int>ma;
       for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];
            ma[a[i]]++;
        }
        for(int i=0;i<n;i++)
        {
            sum-=a[i];
            ma[a[i]]--;
            if(sum%2==0&&ma[sum/2])
            {
                bool donnot=0;
                fi(j,0,n)
                {
                    if(j==i);
                    else if(donnot==0&&a[j]==sum/2)
                    {
                        donnot=1;
                    }
                    else
                    {
                        printf("%lld ",a[j]);
                    }
                }
                puts("");
                goto loop;
            }
            sum+=a[i];
            ma[a[i]]++;
        }
        puts("-1");
        loop:;
    }
    return 0;
}                      