Codeforces Round 49 C. Minimum Value Rectangle

 

1. problem 链接:http://codeforces.com/contest/1027/problem/C

                                                       C. Minimum Value Rectangle

                                                       time limit per test    2 seconds

                                                  memory limit per test   256 megabytes

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

Copy

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5

output

Copy

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).

1. 先有下面的关系： 下图转自https://blog.csdn.net/Light2Chasers/article/details/81834852
   
   很明显只要a和b足够接近就可以了，刚开始还自己一顿瞎测数据试图找到规律，但是由于偶然性和不可证性导致没有捋清思路，这种题应该用数学公式化简推到一下，就能找到普适性的规律

2. AC code

3. #include
   using namespace std;
   #define ll long long
   int a[1000003];
   int xin[1000003];
   int main()
   {
       int t,n,x;
       cin>>t;
       while(t--)
       {
           scanf("%d",&n);
           int b[10003]= {0};
           int ci[10003]= {0};
           for(int i=0; i=4)
               {
                   ans=a[i];
                   break;
               }
           }
           if(ans==-1)
           {
               int h=0;
               int maxx=-1;
               int maxx1=-1;
   
               for(int i=0; i=2&&b[a[i]]==0)
                   {
                       b[a[i]]=1;
                       xin[h++]=a[i];
                   }
               }
               sort(xin,xin+h);
              double cha=10000000;
               int anss=-1,ansss=-1,lll=0;
               int gg=0;
               for(int i=0; i