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;

void fn(int n, vector<int>& b) {
    n += 2;
    sort(b.begin(), b.end());

    auto s = accumulate(b.begin(), b.end() - 2, (long long) 0);
    if (s == b[n - 2] || s == b[n - 1]) {
        for (auto i = 0; i < n - 2; ++i)
            printf("%d ", b[i]);
        printf("\n");
        return;
    }

    s += b[n - 2];
    for (auto i = 0; i < n - 1; ++i) {
        if (s - b[i] == b[n - 1]) {
            for (auto j = 0; j < n - 1; ++j)
                if(j != i)
                    printf("%d ", b[j]);
            printf("\n");
            return;
        }
    }

    printf("-1\n");
}

int main() {
    int t;
    cin >> t;

    for (auto i = 0; i < t; ++i) {
        int n;
        cin >> n;

        vector<int> b(n + 2);
        for (auto k = 0; k < n + 2; ++k)
            cin >> b[k];
        fn(n, b);
    }
}