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;

long double PI=3.14159265358979323846;

long long int nC2(long long int num) {
    int num1 = num-1;
    if(num%2)
        num1 /= 2;
    else
        num /= 2;
    num *= num1;
    return num;
}

long long int fact(long long int num, long long int mod) {
    long long int ans=1;
    for(int i=2; i<=num; i++) {
        ans *= i;
        ans = ans % mod;
    }
    return ans;
}

int main() {
    
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif

    long long int t; cin >> t;
    while(t--) {
        long long int n, sum=0; cin >> n;
        n += 2;
        vector <long long int> vec(n);
        for(long long int i=0; i<n; i++) {
            cin >> vec[i];
            sum += vec[i];
        }
        sort(vec.begin(), vec.end());
        long long int last = vec[n-1], last_second = vec[n-2];
        long long int last_left = sum - last, last_second_left = sum - last_second;
        long long int i_last=-1, i_last_second=-1;
        for(long long int i=0; i<n; i++) {
            if(i == n-1)
                continue;
            if(last_left - vec[i] == last) {
                i_last = i;
                break;
            }
        }
        if(i_last == -1) {
            for(int i=0; i<n; i++) {
                if(i == n-2)
                    continue;
                if(last_second_left - vec[i] == last_second) {
                    i_last_second = i;
                    break;
                }
            }
        }
        vector <long long int> ans(n);
        if(i_last != -1) {
            for(long long int i=0; i<n; i++) {
                if(i == i_last || i == n-1)
                    continue;
                cout << vec[i] << ' ';
            }
        } else if(i_last_second != -1) {
            for(long long int i=0; i<n; i++) {
                if(i == i_last_second || i == n-2)
                    continue;
                cout << vec[i] << ' ';
            }
        } else {
            cout << "-1";
        }
        cout << endl;
    }

    return 0;
}