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<iostream>
#include<vector>
#include <unordered_map>  
#include <algorithm>
#include <functional>
#include <map>
#include <string>
#include <sstream>
#include <cmath>
#include <float.h>
using namespace std;



#define N 100000

long long int binsearch_lower_bound(vector<long long int>& vec, int val) {
	long long int low = 0;            // 範囲の下限
	long long int high = vec.size();  // 範囲の上限。解はあれば [low, high] の中に存在する
	while (low < high) {
		long long int mid = (low + high) / 2;  // 中心点（この実装ではオーバーフローしうる。
									 // 　厳密には mid=low+(high-low)/2 とすべき）
		long long int midval = vec[mid];       // 中心での値

									 // 解の範囲を狭める
		if (midval < val) low = mid + 1;
		else high = mid;
	}
	return low;
}

int main(){
	long long int n, tmp;
	double middle;
	vector<long long int> a;
	cin >> n;
	for (long long int i = 0; i < n; ++i) {
		cin >> tmp;
		a.push_back(tmp);
	}
	sort(a.begin(), a.end());
	if (n == 2) {
		cout << a[1] << " " << a[0];
		return 0;
	}
	double d = (double)a[n - 1];
	middle = d / 2;
	tmp = binsearch_lower_bound(a, middle);
	double b = (double)a[tmp], c = (double)a[tmp - 1];
	if (fabs(b - middle) < fabs(c - middle)) {
		cout << a[n - 1] << " " << a[tmp];
	}
	else {
		cout << a[n - 1] << " " << a[tmp - 1];
	}
	return 0;
}