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>
#include <iomanip>
using namespace std;
#define ll long long
#define db double
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define ppb pop_back
#define mpll map<ll, ll>
#define mcll map<char,ll>
#define msll map<string,ll>
#define mpll_p map<ll, pair<ll, ll>>
#define mp_pll map<pair<ll, ll>, ll>
#define vll vector<ll>
#define vpll vector<pair<ll, ll>>
#define vtll vector<tuple<ll,ll,ll>> 
#define vs vector<string>
#define vc vector<char>
#define sll set<ll>
#define spll set<pair<ll, ll>>
#define nope string::npos
#define endl "\n"
ll mod=1e9+7;
void fast(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool primecheck(ll n){
  for(ll i=2;i<=sqrt(n);i++){
    if(n%i==0){
      return 0;
    }
  }
  return 1;
}
ll power(ll x,ll y,ll mod) { 
    ll res = 1; // Initialize result 
  
    while (y > 0) { 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res =( res * x)%(mod); 
  
        // n must be even now 
        y = y >> 1; // y = y/2 
        x = (x * x)%mod; // Change x to x^2 
    } 
    return res%mod; 
}

int main(){
  fast();
  ll t; cin>>t;
  while (t--)
  {
    ll n; cin>>n;
    vll b; ll sum=0;
    for(ll i=0;i<n+2;i++){
      ll x; cin>>x; sum+=x;
      b.pb(x);
    }
    sort(b.begin(),b.end());
    ll f=-1; 
    for(ll i=0;i<n+2;i++){
      if(i!=n+1){
        sum-=b[i];
        if(sum==2*b.back()){
          f=i; break;
        }
        sum+=b[i];
      }
      else{
        sum-=b[i];
        ll tm=b[i]; b.ppb();
        if(sum==2*b.back()){
          f=i; b.pb(tm); break;
        }
        else{
          sum+=b[i]; b.pb(tm);
        }
      }
    }
    if(f==-1){cout<<-1<<endl;}
    else{
      if(f==n+1){b.ppb(); b.ppb();}
      else{b.erase(b.begin()+f); b.ppb();}
      for(auto x:b){cout<<x<<' ';}
      cout<<endl;
    }
  }
}