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>
#define ll long long
using namespace std ;
 int n , m ;
long long x;

    int main() {
        int t ;
        scanf("%d" ,&t) ;
         while (t--) {
        scanf("%d" ,&n) ;
        vector<long long > vec ;
        long long su = 0 ;
        for (int i = 0 ; i < n  + 2 ; i++ ){
            scanf("%I64d" , &x);
            vec.push_back(x) ;
            su+= vec[i] ;
         }

    sort (vec.begin () , vec.end () );
    int k = n + 1;
    bool ok = false;
     su -= vec[k] ;
    // try sum is k "
  int e =  -1 ;
    for (int i = 0 ; i <= n ; i ++ ) {
        if (su - vec[i] == vec[k] ) {
           e = i ;
            ok = true;
            break ;
        }
    }
     if (ok ){
        for (int i = 0 ; i <=n ; i++) {
           if (i != e )printf ("%I64d " , vec[i] ) ;

        }
        printf("\n");
        continue ;
    }

    ok = (su - vec[k-1] == vec[k-1])    ;
      if (ok ){
        for (int i = 0 ; i <n ; i++) {
           printf ("%I64d " , vec[i] ) ;

        }
        printf("\n");
    }else printf("-1\n");

    }
    }