Problem:
Score : 400 points

Problem Statement

Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order.
From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized.
If there are multiple pairs that maximize the value, any of them is accepted.

Constraints

2 \leq n \leq 10^5

0 \leq a_i \leq 10^9

a_1,a_2,...,a_n are pairwise distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print a_i and a_j that you selected, with a space in between.

Sample Input 1

5
6 9 4 2 11

Sample Output 1

11 6

\rm{comb}(a_i,a_j) for each possible selection is as follows:

\rm{comb}(4,2)=6

\rm{comb}(6,2)=15

\rm{comb}(6,4)=15

\rm{comb}(9,2)=36

\rm{comb}(9,4)=126

\rm{comb}(9,6)=84

\rm{comb}(11,2)=55

\rm{comb}(11,4)=330

\rm{comb}(11,6)=462

\rm{comb}(11,9)=55

Thus, we should print 11 and 6.

Sample Input 2

2
100 0

Sample Output 2

100 0

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
class Program
{
    static void Main(string[] args)
    {
            // 入力
            string[] sS = Console.ReadLine().Split(' ');
            int iN = int.Parse(sS[0]);
            sS = Console.ReadLine().Split(' ');
            List<int> lstA = new List<int>();
            for (int i = 0; i < sS.Length; i++) { lstA.Add(int.Parse(sS[i])); }

            System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();
            sw.Start();

            // 解析
            lstA.Sort();                                                                                // 一応昇順
            int iMax = lstA[lstA.Count - 1];                                                            // nCrのnはすぐ判明
            int iMid = iMax / 2;                                                                        // nCrのrはこの値が最大だか数列内存在するかは不明
            if (lstA.Count == 2) { iMid = lstA[0]; iMax = lstA[1]; }                                    // 2つしかない場合 固定
            else                                                                                        // 3つ以上  数列内でrに近い値をどうにか取得
            {
                List<ClsPot> lstPot = new List<ClsPot>();
                for (int i = 0; i < lstA.Count - 1; i++)                                                // lstA.Count - 1 まで。最後のはiMaxで確定済み
                {
                    lstPot.Add(new ClsPot() { iX = lstA[i], iD = Math.Abs(iMid - lstA[i]) });           // 値とrまでの差
                }
                lstPot.Sort(delegate (ClsPot mca1, ClsPot mca2) { return (int)(mca1.iD - mca2.iD); });  // iDの昇順
                iMid = lstPot[0].iX;                                                                    // rまでの差が一番小さいの
                if (lstPot[0].iD == lstPot[1].iD) { iMid = lstPot[0].iX > lstPot[1].iX ? lstPot[0].iX : lstPot[1].iX; } // 2番目に小さいのも同じ差ならXの大きいの
            }

            // 出力
            Console.WriteLine("{0} {1}", iMax.ToString().Trim(), iMid.ToString().Trim());

            sw.Stop();
#if DEBUG
            Console.WriteLine();
            Console.WriteLine("{0}", sw.Elapsed.ToString());
            Console.WriteLine("続行するには何かキーを押してください．．．");
            Console.ReadKey();
#endif
    }

        // 点位置情報
        private class ClsPot
        {
            public int iX { get; set; } = -1;                        // 値
            public int iD { get; set; } = -1;                        // rとの差
        }

}