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;
#define int long long
#define ll long long
#define ld long double
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using P=pair<long long,long long>;
#define rep(i,n) for(long long i=0; i<(long long)n; i++)
#define FOR(i,a,b) for(long long i=a; i<b; i++)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((long long)(x).size())
#define COUT(x) cout << x << endl
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define vint vector<int>
#define vvint vector<vector<int>>
#define vstr vector<string>
#define vp vector<pair<int,int>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define SUM(x) accumulate(x.begin(),x.end(),0)
#define MAX(x) *max_element(x.begin(),x.end())
#define MIN(x) *min_element(x.begin(),x.end())
#define couty cout<<"Yes"<<endl
#define coutn cout<<"No"<<endl
#define coutY cout<<"YES"<<endl
#define coutN cout<<"NO"<<endl
#define yn(x) cout<<(x?"Yes":"No")<<endl
#define YN(x) cout<<(x?"YES":"NO")<<endl
long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}
long long lcm(long long a, long long b){return a*b/gcd(a,b);}
const long long dx[4]={1,0,-1,0};
const long long dy[4]={0,1,0,-1};
const long long INF = 1e12;
const long long MOD = 1e9+7;

signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout<<fixed<<setprecision(15);

    int n; cin >> n;
    vint a(n);
    rep(i,n) cin >> a[i];
    sort(all(a));
    int p,q;
    //pCqを求める
    //pはaの中で最大の数,qはp/2に近いほどよい
    p=a[n-1];

    int max=-1*INF;
    ld half=(ld)p/2.0;
    ld min=INF;
    rep(i,n-1){    
        if(abs(half-a[i])<min){
            chmin(min,abs(half-a[i]));
            q=a[i];
        }
    }

    cout<<p<<' '<<q<<endl;
    
}