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<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<pdd> vdd;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define mem(a,b) memset(a, b, sizeof(a) )
#define all(x) (x).begin(),(x).end()
#define INF 1000000000000
#define MOD 1000000007
#define PB push_back
#define MP make_pair
#define F first
#define S second

inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

int ctoi(char c) {
	if (c >= '0' && c <= '9') {
		return c - '0';
	}
	return 0;
}

int main(){
	int n;cin>>n;
	vl a(n);
	rep(i,n){
		cin>>a[i];
	}
	sort(all(a));
	if(a[n-1]%2==1){
		int l=a[n-1]/2,r=a[n-1]/2+1,ans=1e9;
		rep(i,n){
			if(abs(ans-l)>abs(l-a[i])){
				ans=a[i];
			}
			if(abs(ans-r)>abs(r-a[i])){
				ans=a[i];
			}
		}
		cout<<a[n-1]<<" "<<ans<<endl;
	}
	else{
		int m=a[n-1]/2,ans=1e9;
		rep(i,n){
			if(abs(ans-m)>abs(m-a[i])){
				ans=a[i];
			}
		}
		cout<<a[n-1]<<" "<<ans<<endl;
	}
}