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>
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define FORR(i,a,b) for (int i=(a);i>=(b);i--)
#define pb push_back
#define pcnt __builtin_popcount
#define show(x) cout<<#x<<" = "<<x<<endl;
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define fi first
#define se second
#define rng(a) (a.begin()),(a.end())
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define sz(x) (int)(x).size()
#define mp make_pair

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef set<int> si;
typedef pair<ll,ll> pll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<pll> vpll;
typedef set<ll> sl;
typedef __int128_t lll;
typedef pair<lll,lll> plll;
typedef vector<lll> vlll;
template<typename T>string join(vector<T>&v)
{stringstream s;FOR(i,0,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename A, size_t N, typename T>void Fill(A (&array)[N], const T&v)
{fill((T*)array,(T*)(array+N),v);}
lll gcd(lll a,lll b,lll &x,lll &y){if(!b){x=1;y=0;return a;}lll d=gcd(b,a%b,y,x);y-=a/b*x;return d;}
ll gcd(ll a,ll b){lll x=0,y=0; return gcd(a, b, x, y);}
ll modpow(lll a,lll n,ll m){if(a==0)return a;lll p=1;for(;n>0;n/=2,a=a*a%m)if(n&1)p=p*a%m;return(ll)p;}
void dout(double d){printf("%.12f\n",d);}

const int iinf = 1e9;
const ll linf = 1e18;
const int mod = 1e9+7;
const double pi = acos(-1);
const double eps = 1e-10;
int n, a[100000];
main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n;
  FOR(i, 0, n) cin >> a[i];
  sort(a, a + n);
  int j = -1, ans = -1;
  FOR(i, 0, n-1){
    if(min(a[i], a[n-1] - a[i]) > j){
      ans = a[i];
      j = min(a[i], a[n-1] - a[i]);
    }
  }
  cout << a[n-1] << " " << ans << endl;
  return 0;
}