You have nn sticks of the given lengths.

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let SS be the area of the rectangle and PP be the perimeter of the rectangle.

The chosen rectangle should have the value P2SP2S minimal possible. The value is taken without any rounding.

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input
The first line contains a single integer TT (T≥1T≥1) — the number of lists of sticks in the testcase.

Then 2T2T lines follow — lines (2i−1)(2i−1) and 2i2i of them describe the ii-th list. The first line of the pair contains a single integer nn (4≤n≤1064≤n≤106) — the number of sticks in the ii-th list. The second line of the pair contains nn integers a1,a2,…,ana1,a2,…,an (1≤aj≤1041≤aj≤104) — lengths of the sticks in the ii-th list.

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all TT lists doesn’t exceed 106106 in each testcase.

Output
Print TT lines. The ii-th line should contain the answer to the ii-th list of the input. That is the lengths of the four sticks you choose from the ii-th list, so that they form a rectangle and the value P2SP2S of this rectangle is minimal possible. You can print these four lengths in arbitrary order.

If there are multiple answers, print any of them.

Example
Input
3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5
Output
2 7 7 2
2 2 1 1
5 5 5 5
Note
There is only one way to choose four sticks in the first list, they form a rectangle with sides 22 and 77, its area is 2⋅7=142⋅7=14, perimeter is 2(2+7)=182(2+7)=18. 18214≈23.14318214≈23.143.

The second list contains subsets of four sticks that can form rectangles with sides (1,2)(1,2), (2,8)(2,8) and (1,8)(1,8). Their values are 622=18622=18, 20216=2520216=25 and 1828=40.51828=40.5, respectively. The minimal one of them is the rectangle (1,2)(1,2).

You can choose any four of the 55 given sticks from the third list, they will form a square with side 55, which is still a rectangle with sides (5,5)(5,5).
為數不多的數論題，主要是因為數學學的實在不怎麼樣。
這個題的意思就是在一堆數中找出符合條件的四個數構成一個矩形。並且這個矩形還要滿足周長的平方除以面積的值最小。簡單推理一下：

以上的就是推導過程。這樣我們就發現，在a一定的情況下，我們只要找大於a的最小的b就可以了。程式碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;

const int maxx=1e6+10;
ll a[maxx];
ll b[maxx];
ll n;

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		cin>>n; 
		for(ll i=0;i<n;i++)
		cin>>a[i];
		sort(a,a+n);//將元素由大到小排好序，這樣能夠構成矩形的元素就在一起了。
		int cnt=0;
		for(ll i=1;i<n;)//將能夠構成矩形的元素存到一個數組裡，方便檢查。
		{
			if(a[i]==a[i-1])
			{
				b[cnt++]=a[i];
				i+=2;
			}
			else i+=1;//在這個地方wa了一次。
		}
		double ans=inf;
		ll x;
		ll y;
		for(ll i=1;i<cnt;i++)
		{
			double ant=double((double)b[i]/b[i-1]+(double)b[i-1]/b[i]);//強制轉換為double型別。
			if(ans>ant)
			{
				ans=ant;
				x=b[i-1];
				y=b[i];
			}
		}
		printf("%lld %lld %lld %lld\n",x,x,y,y);
	}
	return 0;
}

努力加油a啊，(^o)/~