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>
const double pi = acos(-1.0);
using namespace std;
//using namespace __gnu_pbds;
#define   endl    '\n'
#define   sl(n)     scanf("%lld",&n)
#define   mp      make_pair
#define   pb      push_back
#define   ppb     pop_back
#define   fi      first
#define   se      second
#define   ll      long long
#define   ull     unsigned long long
#define   ld      long double
#define   pii     pair<int, int>
#define   f(i,a,b)  for(ll i = (ll)(a); i < (ll)(b); i++)
#define   rf(i,a,b)   for(ll i = (ll)(a); i > (ll)(b); i--)
#define   ms(a,b)   memset((a),(b),sizeof(a))
#define   abs(x)    ((x<0)?(-(x)):(x))
#define   MAX     200005
#define   inf     LLONG_MAX
#define   ninf    LLONG_MIN
#define   MIN     INT_MIN
#define   yeet    return 0;
#define tihs if(fopen("input.txt","r"))freopen("input.txt", "r", stdin),freopen("output.txt", "w", stdout);

#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// Use cout.flush() for interactive problems.
// Use this for increased stack size: g++ -o a.exe -Wl,--stack=256000000 -O2 source.cpp
inline long long  MAX2(long long  a, long long  b) {return (a) > (b) ? (a) : (b);}
inline long long  MAX3(long long  a, long long  b, long long  c) {return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));}
inline long long  MIN2(long long  a, long long  b) {return (a) < (b) ? (a) : (b);}
inline long long  MIN3(long long  a, long long b, long long c) {return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));}

//typedef
typedef long int int32;
typedef unsigned long int uint32;
typedef long long int int64;
typedef unsigned long long int  uint64;


int mod = 1e9 + 7 ;
int64_t extGcd(int64_t a, int64_t b, int64_t& x, int64_t& y) {if (!a) {x = 0; y = 1; return b;} int64_t x1, y1; int64_t d = extGcd(b % a, a, x1, y1); x = y1 - (b / a) * x1; y = x1; return d;}
inline ll addmod(ll a, ll b) {a = a % mod + b % mod; if (a > mod)a %= mod; return a;}
inline ll submod(ll a, ll b) {a = a % mod - b % mod; if (a < 0)a += mod; return a;}
inline ll mulmod(ll a, ll b) {return (a % mod * b % mod) % mod;}

int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
inline ll exp(ll a, ll b) {if (a == 0)return 0ll; ll r = 1LL; while (b > 0) {if (b & 1) {r = r * (a % mod); r = (r + mod) % mod;} b /= 2; a = (a % mod) * (a % mod); a = (a + mod) % mod;} return (r + mod) % mod;}
ll gcd(ll a, ll b) {if (b == 0)return a; if (a == 0)return b; return gcd(b, a % b);}
uint32 setbits(ll n) {uint32 count = 0; while (n) {n &= (n - 1); count++;} return count; }
// ll f[MAX];
// ll iv[MAX];
// ll C(ll n, ll r) {
// 	if (n < r)return 0;
// 	ll ans = (f[n] % mod * iv[r] % mod) % mod;
// 	ans = (ans % mod * iv[n - r] % mod) % mod;
// 	return ans % mod;
// }
// void init() {
// 	f[0] = 1;
// 	iv[0] = 1;
// 	f(i, 1, MAX)f[i] = (i * f[i - 1]) % mod;
// 	iv[MAX - 1] = exp(f[MAX - 1], mod - 2);


// 	for (int i = MAX - 2; i >= 0; --i)iv[i] = (iv[i + 1] * (i + 1)) % mod;

// }
////****************************************************************************************************************************************************************************************************************////

int main() {
	tihs;
	fast_io;
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		vector<ll>g1(n + 2);
		for (int i = 0; i < n + 2; i++)cin >> g1[i];
		sort(g1.begin(), g1.end());
		ll sum1 = 0;
		int m = n + 2;
		f(i, 0, m)sum1 += g1[i];
		ll rest_sum1 = sum1 - g1[m - 1];
		int extra_id = -1;
		for (int i = 0; i < m - 1; i++) {
			ll temp = rest_sum1 - g1[i];
			if (temp == g1[m - 1]) {
				extra_id = i;
				break;
			}

		}
		// cout << rest_sum1 << " " << sum1 << endl;
		// cout << extra_id << endl;
		if (extra_id != -1) {
			vector<ll>ans;
			for (int i = 0; i < m - 1; i++) {
				if (extra_id == i)continue;
				else ans.pb(g1[i]);
			}
			for (ll i : ans)cout << i << " ";
			cout << endl;
			continue;
		}
		rest_sum1 = sum1 - g1[m - 2] - g1[m - 1];
		// cout << rest_sum1 << endl;
		ll po = 0;
		for (int i = 0; i < n; i++)po += g1[i];
		if (po == g1[m - 2])extra_id = m - 1;

		if (extra_id != -1) {
			vector<ll>ans;
			for (int i = 0; i < m ; i++) {
				if (extra_id == i or i == m - 2)continue;
				else ans.pb(g1[i]);
			}
			for (ll i : ans)cout << i << " ";
		}
		else {cout << -1 ;}
		cout << endl;




	}






	return 0;
}