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;

typedef long long int ll;

auto io_speed = []() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    return 0;
}();

ll pow_mod(ll a, ll e, ll mod) {
    ll res = 1;
    while(e) {
        if(e&1) res = res * a%mod;
        e>>=1;
        a=a*a%mod;
    }
    return res;
}

ll inv_mod(ll b, ll m) {
    ll a=m, x=0, y=1, r, q;
    while(b) {
        r = a%b;
        q = a/b;
        a = b;
        b = r;
        r = x;
        x = y;
        y = r-q*y;
    }
    x+=m;
    return x%m;
}

void solve() {
    int n;
    cin >> n;
    n+=2;
    vector<ll> arr(n);
    for(auto &x : arr) cin >> x;

    ll sum = accumulate(arr.begin(), arr.end(), 0ll);
    map<ll, vector<int>> vals;
    for(int i = 0; i < n; i++) {
        vals[arr[i]].push_back(i);
    }

    bool ok = false;
    vector<ll> res;
    ll a = -1, b = -1;
    for(int i = 0; i < n; i++) {
        ll tmp = sum - arr[i];
        if(tmp%2 == 0) {
            tmp/=2;
            auto itr = vals.find(tmp);
            if(itr != vals.end()) {
                if(tmp == arr[i]) {
                    if(itr->second.size() > 1) {
                        ok = true;
                        a = b = tmp;
                        break;
                    }
                } else {
                    ok = true;
                    b = arr[i];
                    a = tmp;
                    break;
                }
            }
        }
    }
    
    if(ok) {
        vals[a].pop_back();
        vals[b].pop_back();
        for(auto x : vals) {
            for(auto y : x.second) {
                cout << arr[y] << " ";
            }
        }
        cout << endl;
    } else cout << -1 << endl;
}

int main() {
    int t = 1;
    cin >> t;
    for(int i = 1; i <=t; i++) {
        // cout << "Case #"<<i<<": ";
        solve();
    }
}