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;

using ll = long long;
using ld = long double;

#define REP(i,a)       for(ll i = 0; i < a; ++i)
#define RREP(i,a)      for(ll i = a-1; i >= 0; --i)
#define REPA(i,a,b)  for(ll i = (a); i < (b); ++i)

#define all(x) (x).begin(), (x).end()
#define pb push_back
#define F first
#define S second
#define bitcount __builtin_popcount
#define YesNo(condition) cout << (condition ? "Yes" : "No") << "\n"
#define YESNO(condition) cout << (condition ? "YES" : "NO") << "\n"

const ld PI = acos((ld)-1);
const ll mod = 1e9+7;
// const ll mod = 998244353;

// int lcm(int a, int b){
//     return a/(__gcd(a, b))*b;
// }

// ll binpow(ll a, ll b) {
//     a %= mod;
//     ll res = 1;
//     while (b > 0) {
//         if (b & 1) res = res * a % mod;
//         a = a * a % mod;
//         b >>= 1;
//     }
//     return res;
// }

// bool isPrime(ll x){
//     if(x==2 || x==3) return 1;
//     if(x%2==0) return 0;
//     for(ll i=3; i<=sqrt(x); i += 2)
//         if(x%i==0) return 0;
//     return 1;
// }

// ll primesize = 1e8+5;
// vector<bool> prime(primesize, true);
// void makeSieve(ll num = primesize){
//     for(ll i=4; i<= num; i += 2) 
//         prime[i] = false;
//     for(ll p = 3; p*p <= num; p += 2) 
//         if(prime[p])
//             for(ll i = p*p; i<=num; i += p) 
//                 prime[i] = false;
// }

ll n, m;
string s;

void solve(){
    cin >> n;
    int a[n+2];
    REP(i,n+2) cin >> a[i];
    sort(a,a+n+2);
    
    bool possible = false, done = false;

    ll sum = accumulate(a,a+n+1,0ll);
    for(int i=0; i<n+1; ++i) {
        if(sum == a[n+1] + a[i]) {
            a[i] = 0;
            a[n+1] = 0;
            done = true;
            break;
        }
    }


    if(!done) possible = (sum-2*a[n] == 0);

    if(done || possible) {
        if(!done) a[n] = a[n+1] = 0;
        REP(i,n+2) if(a[i]) cout << a[i] << " ";
        cout << "\n";
    } else {
        cout << "-1\n";
    }
}

#pragma region
int main(){
    ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
    cout << setprecision(12) << fixed;

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

    ll tc=1;
    cin >> tc;
    while(tc--) 
    solve();
}
#pragma endregion