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: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using static System.Console;


namespace ABC94_4
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = int.Parse(ReadLine());
            string[] s = ReadLine().Split(' ');
            int[] a = new int[n];
            int[] b = new int[n];
            int ma = 0;
            for (int i = 0; i < n; i++)
            {
                a[i] = int.Parse(s[i]);
                b[i] = int.Parse(s[i]);
                ma = Math.Max(ma, a[i]);
            }

            Array.Sort(b);
            int[] m = new int[3];
            if (n % 2 == 0)
            {
                m[0] = b[n / 2 - 1];
                m[1] = b[n / 2];
                m[2] = b[n / 2];
            }
            else
            {
                m[0] = b[(n-1) / 2];
                m[1] = b[(n-1) / 2 + 1];
                m[2] = b[(n - 1) / 2 + 2];
            }
            long mb = 0;
            int ans = 0;
            for (int i = 0; i < 2; i++)
            {
                long num = 1;
                for (int j = 1; j <= m[i]; j++)
                {
                    num = num * (ma - j + 1) / j;
                }
                if (num > mb)
                {
                    mb = num;
                    ans = m[i];
                }
            }

            WriteLine(ma + " " + ans);
        }
    }
}