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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>
#include <algorithm>
#include <vector>

int main() {
  int n;
  scanf("%d", &n);

  std::vector<int> a(n);
  for (int i=0; i<n; ++i)
    scanf("%d", &a[i]);

  std::sort(a.begin(), a.end());

  if (n == 2)
    return !printf("%d %d\n", a[0], a[1]);

  int N=a.back();
  int j=std::lower_bound(a.begin(), a.end()-1, N/2)-a.begin();

  if (j == N/2)
    return !printf("%d %d\n", N, a[j]);

  if (j > 0 && a[j]-N/2 > N/2-a[j-1])
    --j;

  printf("%d %d\n", N, a[j]);
}