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 fast  ios_base::sync_with_stdio(false);  cin.tie(nullptr), cout.tie(nullptr)
#define FIO  freopen("input.txt","r",stdin);  freopen("output.txt","w",stdout);
#define gcd __gcd
#define pi acos(-1)
#define sp " "
#define LL long long
#define mod (long long)(1000000000 + 7)
#define SET(n) fixed << setprecision(n)
int ioo = INT_MAX;
LL Loo = LONG_LONG_MAX;

LL power(LL fa, LL n)
{
    LL k = 1;
    while(n)
    {
        if(n % 2) k = (k * fa) % mod;
        fa = (fa * fa) % mod;
        n /= 2;
    }
    return k;
}

int n;
LL a[200005];

int main() {
    fast;

    int q; cin >> q;
    while(q--)
    {
        cin >> n;
        LL sum = 0;
        for(int i = 1;i <= n+2; i++)
        {
            cin >> a[i];
            sum += a[i];
        }
        sort(a + 1 , a + n + 3);

        int x = -1 , n1 = -1;
        for(int i = 1;i <= n+1; i++)
        {
            if((sum - a[i]) - a[n+2] == a[n+2])
            {
                x = i;
                n1 = n+2;
                break;
            }
        }
        if((sum - a[n+2]) - a[n+1] == a[n+1]) x = n+2 , n1 = n+1;

        if(x == -1) cout << -1 << endl;
        else
        {
            for(int i = 1;i <= n+2; i++) if(i != x && i != n1) cout << a[i] << sp;
            cout << endl;
        }
    }
    return 0;
}