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: Runtime Error)

n = gets.chomp.to_i
as = gets.chomp.split.map(&:to_i)

# 階乗の計算
def factorial(n)
  if n == 1
    return 1
  elsif n == 0
    return 1
  else
    return n*factorial(n-1)
  end
end

# 組み合わせの計算
def comb(n,r)
  return factorial(n)/factorial(n-r)/factorial(r)
end

# ここまでOK

answer = 0
tmp = 0
ary = [0,0]

for i in 0..n-2 do
  for j in i+1..n-1 do
    if as[i] < as[j] then
      tmp = comb(as[j],as[i])
    else
      tmp = comb(as[i],as[j])
    end
    if answer < tmp then
      answer = tmp
      ary = [as[i],as[j]]
    end
  end
end

ary.sort! {|a, b| b<=> a }
puts(ary[0].to_s+" "+ary[1].to_s)