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>

typedef int i32;
typedef long long i64;
typedef unsigned int u32;
typedef unsigned long long u64;


template<typename First, typename Second, typename Third>
struct triple {
   First first;
   Second second;
   Third third;
};


#define debug(x)  cerr<<x<<"\n";
#define loop(i,n) for(i32 i = 0; i < n; ++i)
#define pi  pair<int,int>
#define ti  tuple<int,int,int>
#define mt  make_tuple
#define mk  make_pair
#define vi  vector<int>
#define vll vector<i64>
#define fastInOut ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define pr(a) for(auto x:a) cout<<x<<" ";
#define prmap(m) for(auto x:m) cout<<x.first<<" "<<x.second<<"\n";
using namespace std;

int main() 
{ 
	i32 n,k;
    cin>>n;
    vector<double> v(n);
    //vector<int> temp(n);
    for (int i = 0; i < n; i++) {
        cin >> v[i];
    }
	//temp=v;
    sort(v.begin(), v.end());
	double mn=v[0];
    for (int i = 1; i < n; i++) {
        if (abs( mn - v[v.size()-1] / 2.0) > abs(v[i]-v[v.size()-1] / 2.0)) {
            mn = v[i];
        }
    }
    cout<<v[v.size()-1]<<" "<<mn;
	
    return 0; 
}