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"
using namespace std;
#define ll long long int
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll a, b, t, n, k;
  string s;
  cin >> t;
  for (; t; t--) {
    cin >> n;
    vector <ll> v;
    ll totalSum = 0;
    for (ll i = 1; i <= n + 2; i++) {
      cin >> k;
      v.push_back(k);
      totalSum += k;
    }
    sort(v.begin(), v.end());
    ll idx = INT_MIN;
    for (ll i = 0; i <= n; i++) {
      if (totalSum - v[i] == 2 * v.back()) {
        idx = i;
        break;
      }
    }
    for (ll i = 0; i <= n + 1; i++) {
      if (i == n) continue;
      else {
        if (totalSum - v[i] == 2 * v[n]) {
          idx = i;
          break;
        }
      }
    }
    if (idx == INT_MIN) {
        cout << -1 << endl;
        continue;
    }
    vector <ll> printt;
    for (ll i = 0; i <= n + 1; i++) {
      if (i != idx) printt.push_back(v[i]);
    }
    printt.pop_back();
    for (ll i : printt) cout << i << " ";
    cout << endl;
  }
  return 0;
}