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;
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REPR(i,n) for(int i=n; i>-1; --i)
#define ALL(a) (a).begin(),(a).end()
#define FILL(a,n,x); REP(i,(n)){ (a)[i]=(x); }
#define CINA(a,n); REP(i,(n)){ cin >> (a)[i]; }
#define FILL2(a,n,m,x); REP(i,(n)){ REP(j,(m)){(a)[i][j]=(x);} }
#define CINA2(a,n,m); REP(i,(n)){ REP(j,(m)){cin >> (a)[i][j];} }
#define Liny "Yes\n"
#define Linn "No\n"
#define LINY "YES\n"
#define LINN "NO\n"

//cout << setfill('0') << right << setw(4) << 12; // "0012"
int keta(ll x){ if(x<10){return 1;} else{return keta(x/10) + 1;}}
int keta_wa(ll x){ if(x<10){return x;} else{return keta_wa(x/10) + x%10;} }
int ctoi(char c){ return ( (c>='0' && c<='9')? c - '0': 0 );}
int __stoi(string s){ return atoi(s.c_str()); }
ll sum(ll a[],ll N){ return accumulate(a,a+N,0LL);}
ll gcd(ll a,ll b){if(a<b)swap(a,b); return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){if(a<b){swap(a,b);} return a/gcd(a,b)*b;}
template<class T> void chmax(T& a, T b){ if(a<b){a=b;} }
template<class T> void chmin(T& a, T b){ if(a>b){a=b;} }
const ll MOD = 1e9+7;
#define pii pair<int,int>
#define pll pair<ll,ll>

int main(){
  int N; cin>>N;
  ll a[N]; CINA(a,N);
  ll x = -1;
  REP(i,N){
    chmax(x,a[i]);
  }
  ll y = -1;
  ll mid = x/2;
  ll d_mid = x;
  REP(i,N){
    if(x==a[i]){ continue; }
    ll temp = abs(a[i]-mid);
    if(temp<d_mid){
      d_mid = temp;
      y = a[i];
    }
  }
  cout << x << " " << y << "\n";
}