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 <algorithm>
#include <vector>
#include <string>
#include <numeric>
    
using namespace std;    
typedef long long ll;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

int main(){

    int n;
    cin >> n;
    vector<ll> v(n);
    for(int i = 0; i < n; i++){
        cin >> v[i];
    }
    sort(all(v));

    ll minDiff = abs(v.back() - v[0]);
    ll minNum = v[0];
    for(int i = 0; i < n; i++){
        if(minDiff > abs( (v.back()+1 ) / 2 - v[i]) ){
            minDiff = abs( (v.back()+1) / 2 - v[i]);
            minNum = v[i];
        }
    }
    cout << v.back() << " " << minNum << endl;

}