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:
//#pragma GCC optimize("Ofast")

#include <bits/stdc++.h>

using namespace std;

void debug_out() { cerr << endl; }

template<typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
    cerr << "[" << H << "]";
    debug_out(T...);
}

#ifdef dddxxz
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define SZ(s) ((int)s.size())
#define all(x) (x).begin(), (x).end()
#define revall(x) (x).rbegin(), (x).rend()

clock_t startTime;

double getCurrentTime() {
    return (double) (clock() - startTime) / CLOCKS_PER_SEC;
}

typedef long long ll;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
const double eps = 0.00001;
const int MOD = 1e9 + 7;
const int INF = 1000000101;
const long long LLINF = 1223372000000000555;
const int N = 5e5 + 3e2;
const int M = 105;

int a[N], b[N];

void solve(int TC) {
    int n;
    cin >> n;

    ll sum = 0;
    for (int i = 1; i <= n + 2; i++){
        cin >> b[i];
        sum += b[i];
    }

    sort(b + 1, b + n + 3);

    for (int i = 1; i <= n + 2; i++){
        int x = b[i];
        int j = (i == n + 2 ? n + 1 : n + 2);
        if ((sum - x) % 2) continue;

        if (sum - x == b[j] * 2){
            for (int k = 1; k <= n + 2; k++){
                if (k == i || k == j) continue;
                cout << b[k] << ' ';
            }
            cout << endl;
            return;
        }

    }

    cout << -1 << endl;

}

int main() {
    startTime = clock();
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios_base::sync_with_stdio(false);

    bool llololcal = false;
#ifdef dddxxz
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    llololcal = true;
#endif

    int TC = 1;
    cin >> TC;

    for (int test = 1; test <= TC; test++) {
        debug(test);
        //cout << "Case #" << test << ": ";
        solve(test);
    }

    if (llololcal) cerr << endl << "Time: " << int(getCurrentTime() * 1000) << " ms" << endl;

    return 0;
}