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 <iostream>
#include <cstdio>
#include <bitset>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <utility>
#include <string>
#include <algorithm>
#include <cmath>
#include <limits>

using namespace std;
using ll=long long;
using vint=vector<int>;
using pint=pair<int,int>;

const double EPS = 1e-9;
const int inf = numeric_limits<int>::max();

#define rep(i,n) for(int i=0;i<int(n);++i)
#define rrep(i,n) for(int i=int(n-1);i>=0;--i)
#define REP(i,a,b) for(int i=int(a);i<int(b);++i)
#define RREP(i,a,b) for(int i=int(b-1);i>=int(a);--i)
#define SHOW(a) cout << #a << " = " << a << endl
#define ARR(a,n) for(int i=0;i<int(n);++i) cout << #a << "[" << i << "]" << " = " << a[i] << endl

ll combination(unsigned int n, unsigned int r) {
  if ( r * 2 > n ) r = n - r;
  ll dividend = 1;
  ll divisor  = 1;
  for ( unsigned int i = 1; i <= r; ++i ) {
    dividend *= (n-i+1);
    divisor  *= i;
  }
  return dividend / divisor;
}

void solve(){
	int n;
	cin >> n;

	int a[100005];
	rep(i,n) cin >> a[i];
	sort(a, a+n);

	int mm = a[n-1];
	int target = mm / 2;
	auto x = lower_bound(a, a+n, target);
	auto y = upper_bound(a, a+n, target);

	if(abs(mm - *x) > abs(*y - mm)){
		cout << mm << " " << *y << endl;
	}else{
		cout << mm << " " << *x << endl;
	}
}

int main(){
	solve();
	return 0;
}