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: Runtime Error)

#include <stdio.h>

int fact(int x)
{
	int ans=1;
	for(int i=1;i<=x;i++)
	{
		ans*=i;
	}
	return ans;
}

int comb(int x,int y)
{
	if (x>=y)
		return (fact(x)/(fact(x-y)*fact(y)));
	else
		return (fact(y)/(fact(y-x)*fact(x)));

}

int main()
{
	int N;
	scanf("%d",&N);
	int list[N];
	for(int i=0;i<N;i++)
	{
		scanf("%d",&list[i]);
	}
	int max=0;
	int ans[2];
	for(int i=0;i<N-1;i++)
	{
		for(int j=i+1;j<N;j++)
		{
			int num=comb(list[i],list[j]);
			if(num>max)
			{
				max=num;
				if (list[i] >list[j])
				{
					ans[0]=list[i];
					ans[1]=list[j];
				}
				else
				{
					ans[0]=list[j];
					ans[1]=list[i];
				}
			}
		}
	}
	printf("%d %d",ans[0],ans[1]);
	
	return 0;

}