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 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int main(){
	int n;
	cin>>n;
	vector<int>arr(n);
	for(int i=0; i<n; i++) cin>>arr[i];
	sort(arr.begin(),arr.end());
	reverse(arr.begin(),arr.end());
	if(arr[0]%2==0){
		long long mid = arr[0]/2;
		long long diff = numeric_limits<int>::max();
		long long sec;
		for(int i=1; i<arr.size(); i++){
			long long my_diff = abs(mid-arr[i]);
			if(my_diff<diff){
				diff = my_diff;
				sec = arr[i];
			}
		}
		cout<<arr[0]<<" "<<sec;
	}
	else{
		long long mid1 = arr[0]/2+1;
		long long mid2 = arr[0]/2;
		long long diff = numeric_limits<int>::max();
		long long sec=-1;
		for(int i=1; i<arr.size(); i++){
			int diff1 = abs(mid1-arr[i]);
			int diff2 = abs(mid2-arr[i]);
			if(diff1<diff){
				diff = diff1;
				sec = arr[i];
			}
			if(diff2<diff){
				diff = diff2;
				sec = arr[i];
			}
		}
		cout<<arr[0]<<" "<<sec;
	}
}