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 no() {
  cout << "-1" << endl;
}

void solve() {
  int n;
  cin >> n;
  vector<int> b(n + 2);
  for (int &x : b) {
    cin >> x;
  }

  multiset<int> have(b.begin(), b.end());
  long long sum = accumulate(b.begin(), b.end(), 0LL);
  for (int x : b) {
    have.erase(have.find(x));
    sum -= x;
    if (sum % 2 == 0 && sum <= 2000000000 && have.find(sum / 2) != have.end()) {
      have.erase(have.find(sum / 2));
      for (int y : have) {
        cout << y << " ";
      }
      cout << endl;
      return;
    }
    sum += x;
    have.insert(x);
  }
  no();
}

int main() {
  int tests;
  cin >> tests;
  while (tests-- > 0) {
    solve();
  }
  return 0;
}