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;

#define INF 0x3f3f3f3f

typedef unsigned long long ull;
typedef vector<int> vi;


vi solve(vi &nums) {
    ull sum = 0;
    unordered_map<int, int> freqs;

    for (auto x : nums) {
        freqs[x]++;
        sum += x;
    }

    ull cur_sum, skip_a = -1, skip_b = -1;
    for (int i = 0; i < nums.size(); i++) {
        cur_sum = sum - nums[i];
        skip_b = cur_sum - nums[i];
        if (skip_b > 1e9)
            continue;
        
        if ((nums[i] == skip_b && freqs[skip_b] > 1) || (nums[i] != skip_b && freqs[skip_b] > 0))
        {
            skip_a = i;
            break;
        }
    }

    if (skip_a == -1) {
        return vi({-1});
    }

    vi ans;
    bool skipped_b = false;
    for (int i = 0; i < nums.size(); i++)
    {
        if (i == skip_a)
            continue;
        if (nums[i] == skip_b && !skipped_b) {
            skipped_b = true;
            continue;
        }
        ans.push_back(nums[i]);
    }

    return ans;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);
    
    int t, n;
    cin >> t;

    for(int i = 0; i < t; i++) {
        cin >> n;
        vi nums(n + 2);
        for (int i = 0; i < n + 2; i++)
            cin >> nums[i];

        vi ans = solve(nums);
        for (int x : ans)
            cout << x << " ";
        cout << endl;
    }
}