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>
#define int long long
#define hell 1000000007
#define endl "\n"
#define F first
#define S second
#define gcd(a,b) __gcd((a),(b))
#define lcm(a,b) ((a)*(b))/gcd(a,b) 
#define pii pair<int,int>
#define sz(x) (int)x.size()
#define ios ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

using namespace std;

int expo(int base,int exponent,int mod = hell){
    int ans=1;
    base%=mod;
    while(exponent!=0){
        if(exponent&1) ans=(1LL*ans*base)%mod;
        base=(1LL*base*base)%mod;
        exponent>>=1;
    }
    return ans%mod;
}

int inverse(int x){ 
    return expo(x,hell-2,hell);
}

void solve() {
    int n;
    cin>>n;
    vector<int>b(n+2);
    int i;
    int sum=0;
    for(i=0;i<n+2;i++) {
        cin>>b[i];
        sum += b[i];
    }
    multiset<int>has(b.begin(),b.end());
    for(i=0;i<n+2;i++) {
        has.erase(has.find(b[i]));
        sum-=b[i];
        if(sum%2==0 && sum<=2000000000 && has.find(sum/2)!=has.end()) {
            has.erase(has.find(sum/2));
            for(auto it=has.begin();it!=has.end();++it) {
                cout<<*it<<" ";
            }
            cout<<endl;
            return;
        }
        sum+=b[i];
        has.insert(b[i]);
    }
    cout<<-1<<endl;
}
signed main() {
    ios
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    int t=1;
    cin>>t;
    while(t--) {
        solve();
    }
    return 0;
}