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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int main()
{
	int i, n, a[100001];
	scanf("%d", &n);
	for (i = 1; i <= n; i++) scanf("%d", &(a[i]));
	
	int max = 0;
	for (i = 1; i <= n; i++) if (a[i] > max) max = a[i];
	
	int ans = a[1] + 1, min = abs(a[1] * 2 - max) + 1;
	for (i = 1; i <= n; i++) {
		if (a[i] != max && abs(a[i] * 2 - max) < min) {
			min = abs(a[i] * 2 - max);
			ans = a[i];
		}
	}
	
	printf("%d %d\n", max, ans);
	fflush(stdout);
	return 0;
}