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:
#define Fast std::ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pr(x) cout<<#x<<": "<<x<<endl
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
int n;
ll a[maxn],b[maxn];
void output()
{
    for(int i=1;i<=n;++i) printf("%lld%c",a[i],i==n?'\n':' ');
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        for(int i=1;i<=n+2;++i) cin>>b[i];
        sort(b+1,b+1+n+2);
        ll sum=0;
        a[0]=0;
        for(int i=1;i<=n+1;++i) sum+=b[i];
        bool flag=false;
        for(int i=1;i<=n;++i)if((sum-b[i])==b[n+2]){
            for(int j=1;j<=n+1;++j)if(i!=j) a[++a[0]]=b[j];
            flag=true;
            break;
        }
        if(flag){
            output();
            continue;
        }   
        sum-=b[n+1];
        if(sum==b[n+1]||sum==b[n+2]){
            for(int i=1;i<=n;++i) a[++a[0]]=b[i];
            output();
        }
        else puts("-1");
    }
    return 0;
}