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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n=gets.chomp.to_i
array = gets.chomp.split.map(&:to_i)
array.sort!.reverse
counts = []
numbers = []
def kaijo(k)
  start = 1
  while k > 0 do
    start *= k
    k -= 1
  end
  return start
end
def comb(a,b)
  return  kaijo(a) / (kaijo(b) * kaijo(a - b))
end
l = 0
if n >= 3
  n.times do
    halfarray = array.map { |e| (e-array[l]/2).abs  }
    half = halfarray.min + array[l]/2
    numbers << array[l].to_s + " " + half.to_s
    l += 1
  end
  max = counts.max
  puts numbers[-1]
else
  puts array[0].to_s + " " + array[1].to_s
end