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<bits/extc++.h>
#define fr first
#define sc second
using namespace std;
//using namespace __gnu_pbds;
typedef long long ll;
//template<class T>using ordered_set=tree<T,null_type,less<T>,rb_tree_tag,tree_order_statistics_node_update>;
template<class T>void umin(T &a,T b){if(b<a)a=b;}
template<class T>void umax(T &a,T b){if(a<b)a=b;}
int n;
ll a[200010];
int main(){
    int test;
    scanf("%d",&test);
    while(test--){
        scanf("%d",&n);
        n+=2;
        for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
        sort(a+1,a+1+n);
        //If last is sum
        ll sum=0;
        bool ready=false;
        for(int i=1;i<n;i++)sum+=a[i];
        for(int i=1;i<n;i++)if(sum-a[i]==a[n]){
            for(int j=1;j<n;j++)if(i!=j)printf("%d ",a[j]);
            puts("");
            ready=true;
            break;
        }
        if(ready)continue;
        //If last is impostor
        sum-=a[n-1];
        if(sum==a[n-1]){
            ready=true;
            for(int i=1;i<=n-2;i++)printf("%d ",a[i]);
            puts("");
            continue;
        }
        puts("-1");
    }
}