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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <array>
#include <string>
#include <iomanip>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define mp make_pair
#define eb emplace_back
#define F first
#define S second
using namespace std;
using namespace __gnu_pbds;

typedef long long int ll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> ost;

const ll maxN = 1e7;
const ll INF = 1e17;
const ll MOD = 1e9 + 7;

ll xs[] = {-1, 1, 0, 0, -1, 1, 1, -1};
ll ys[] = {0, 0, -1, 1, -1, 1, -1, 1};

ll segtree[maxN << 2];
ll lazy[maxN << 2];
ll A[maxN];

inline ll left(ll x){
	return x << 1;
}

inline ll right(ll x){
	return (x<<1) + 1;
}

void build(ll node, ll start, ll end){
	if(start == end)
		segtree[node] = A[start];
	else{
		ll mid = (start + end)/2;
		build(left(node), start, mid);
		build(right(node), mid+1, end);
		segtree[node] = segtree[left(node)] + segtree[right(node)];
	}
}

void update(ll node, ll start, ll end, ll l, ll r, ll val){
	if(lazy[node] != 0){
		segtree[node] += (end-start+1)*lazy[node];
		if(start != end){
			lazy[left(node)] += lazy[node];
			lazy[right(node)] += lazy[node];
		}
		lazy[node] = 0;
	}
	if(start > end or l > start or r < end)
		return ;
	else if(start >= l and r >= end){
		segtree[node] += val;
		if(start != end){
			lazy[left(node)] += val;
			lazy[right(node)] += val;
		}
	}
	else{
		ll mid=(start + end)/2;
		update(left(node), start, mid, l, r, val);
		update(right(node), mid+1, end, l, r, val);
		segtree[node] = segtree[left(node)] + segtree[right(node)];
	}
}

ll sums(ll node, ll start, ll end, ll l, ll r){
	if(lazy[node] != 0){
		segtree[node] += (end-start+1)*lazy[node];
		if(start != end){
			lazy[left(node)] += lazy[node];
			lazy[right(node)] += lazy[node];
		}
		lazy[node] = 0;
	}
	if(l > end or r < start or start > end)
		return 0;
	if(l <= start and r >= end)
		return segtree[node];
	ll mid = (start+end)/2;
	return sums(left(node), start, mid, l , r) + sums(left(node), mid + 1, end, l , r);
}

int main(){
	cin.sync_with_stdio(false);
	cin.tie(0);
	ll n;
	cin >> n;
	ll a[n];
	for(ll i=0; i<n; i++)
		cin >> a[i];
	sort(a, a+n);
	ll ans = INF, k;
	for(ll i=n-2; i>=0; i--){
		if(ans > abs(a[i]-(a[n-1]/2))){
			k = a[i];
			ans = a[i]-(a[n-1]/2);
		}
	}
	cout << a[n-1] << " " << k;
	return 0;
}