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 Lua (5.3.2) (judge verdict: Accepted)

local mfl = math.floor
local function lower_bound(ary, x)
  if(x <= ary[1]) then return 1 end
  local num = #ary
  if(ary[num] < x) then return num + 1 end
  local min, max = 1, num
  while(1 < max - min) do
    local mid = mfl((min + max) / 2)
    if(ary[mid] < x) then
      min = mid
    else
      max = mid
    end
  end
  return max
end

local n = io.read("*n")
local a = {}
for i = 1, n do a[i] = io.read("*n") end
table.sort(a)
local max = a[n]
table.remove(a)
local idx = lower_bound(a, max / 2)
if(idx == n) then
  idx = n - 1
elseif(1 < idx) then
  if(math.abs(max - 2 * a[idx - 1]) < math.abs(max - 2 * a[idx])) then
    idx = idx - 1
  end
end
print(max .. " " .. a[idx])