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: Accepted)

#include <bits/stdc++.h>
using namespace std;
int n, a[100010];
int main()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	sort(a, a+n);
	int num = a[n-1];
	pair<int, int> mn = { 999999999, 0 };
	int half = (num+1)/2;
	int half2 = half + ((n%2) == 0);
	for (int i = 0; i < n-1; i++)
	{
		mn = min(mn, { abs(a[i]-half), a[i]});
		mn = min(mn, { abs(a[i]-half2), a[i]});
	}
	printf("%d %d\n", num, mn.second);
}