CodeForces - 1027C

Description

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 S be the area of the rectangle and P be the perimeter of the rectangle.

The chosen rectangle should have the value    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 T (T≥1) — the number of lists of sticks in the testcase.

Then 2T lines follow — lines (2i−1) and 2i of them describe the i-th list. The first line of the pair contains a single integer nn () — the number of sticks in the i-th list. The second line of the pair contains n integers a1,a2,…,an() — lengths of the sticks in the i-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 T lists doesn't exceed  in each testcase.

Output

Print T lines. The i-th line should contain the answer to the i-th list of the input. That is the lengths of the four sticks you choose from the i-th list, so that they form a rectangle and the value  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 2 and 7, its area is 2*7=14, perimeter is 2*(2+7)=18. .

The second list contains subsets of four sticks that can form rectangles with sides (1,2), (2,8) and (1,8). Their values are ,  and , respectively. The minimal one of them is the rectangle (1,2).

You can choose any four of the 5 given sticks from the third list, they will form a square with side 5, which is still a rectangle with sides (5,5).

题意：给你n个边长，让你从中找出四个构成一个矩形，使得该矩形的周长的平方除以面积的值最小。

题解：可以证明正方形时最小，因此一定是按照边长排序，相邻的两种边长里一定有最优解。

This is the code

 //注，需要用scanf，表示我用cin时报TLE

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include <iomanip>
#include<list>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x7f7f7f7f      //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了
int a[1000005];
int v[10005];
int out[5];
double ans;
void cmp(int x,int y)
{
    if(a[y+1]!=a[y])//后面连续两个边不一样长
        return ;
    double res=2.0*(a[x]+a[y]);
    res=res*res/(a[x]*a[y]);//计算
    if(res<ans)//判断是否符合
    {
        ans=res;
        out[1]=out[2]=a[x];
        out[3]=out[4]=a[y];
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=1e18;
        memset(v,0,sizeof(v));
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;++i)
        {
            scanf("%d",&a[i]);
            v[a[i]]++;//记录每条边的个数
        }
        for(int i=0;i<n;++i)
            if(v[a[i]]==1)
                a[i]=-1;//将不符合题意的设为-1
        sort(a,a+n);
        int cnt=1;
        for(int i=1;i<n-1;++i)
        {
            if(a[i]==-1)
                continue;
            if(cnt>1)//表示前面有两条相等的边
                cmp(i-1,i);
            if(a[i]==a[i-1])
                cnt++;
            else
                cnt=1;
        }
        for(int i=1;i<5;++i)
            printf("%d ",out[i]);
        printf("\n");
    }
    return 0;
}