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<cstdio>
#include<vector>
#include<set>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
using ll = long long;
const int N = 1e6 + 50;

int T, n;
int b[N];

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        n += 2;
        ll sum = 0;
        int id = -1;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &b[i]);
            sum += b[i];
        }
        sort(b + 1, b + 1 + n);
        if(sum - b[n] - b[n - 1] == b[n - 1]) {
            for(int i = 1; i <= n - 2; ++i)  printf("%d ", b[i]);
            puts("");
            continue;
        }    
        sum -= b[n];
        for(int i = 1; i < n; ++i) {
            if(sum - b[i] == b[n]) {
                id = i;
                break;
            }
        }
        if(id == -1) puts("-1");
        else {
            for(int i = 1; i < n; ++i) {
                if(i != id)  printf("%d ", b[i]);
            }
            puts("");
        }
    }    
}