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 <bits/stdc++.h>
using namespace std;
typedef pair<int,int>P;
 
int n,AA;
vector<int>a;
vector<P>v;
 
signed main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>AA;
		a.push_back(AA);
	}
	sort(a.begin(),a.end());
	for(int i=0;i<n;i++){
		int x=a[i]/2;
		int idx=lower_bound(a.begin(),a.end(),x)-a.begin();
		if(i==n-1){
			if(0){
				if(abs(x-a[idx])>=abs(x-(a[idx-1])))cout<<a[i]<<' '<<a[idx-1]<<endl;
				else cout<<a[i]<<' '<<a[idx]<<endl;
			}
			cout<<a[i]<<' '<<a[idx]<<endl;
		}
	}
	sort(v.begin(),v.end());
 
}