1I_Minimum Value Rectangle

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).

题意：

找出一个矩形数，使他的P^2/S最小。

思路：

我们把这个式子带进去a，b化简可得；只要 $\frac{P^2}{S}=4*(a+b)^2/ab=4*(\frac{a}{b}+\frac{b}{a}+2)$ ,a和b的比值接近1最好。

知道了思路，代码就比较简单了。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f//防止溢出,为什么选择0x3f3f3f3f，有文章可查
using namespace std;

int a[10000];
int b[1000000];//标记数组

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,p=0,i;
        for(i=0;i<=10000;i++)a[i]=0;
        scanf("%d",&n);
        for(i=1; i<=n; i++)
        {
            int m;
            scanf("%d",&m);
            a[m]++;
            if(a[m]==2||a[m]==4)//2 或者4 的时候可以组成一个边
                b[++p]=m;
        }

        sort(b+1,b+1+p);//对b进行排序

        int x,y;
        double minn=inf;

        for(i=1; i<p; i++)
        {
            if(b[i]==b[i+1])
            {
                x=b[i],y=b[i+1];
                break;
            }
            double m=b[i+1]*1.0/b[i];

            if(m<minn)
{minn=m;x=b[i];y=b[i+1];}
        }
        printf("%d %d %d %d\n",x,x,y,y);
    }
    return 0;
}

1I_Minimum Value Rectangle

猜你喜欢