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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 100100100;
const int MOD = (int)1e9 + 7;
const double EPS = 1e-9;
struct Edge {
  ll to;
  ll cost;
};
long long combination(long n, long r, long long mod) {
  // nCrを返す　O(r)
  if (n < r) swap(n, r);
  long long bunbo = 1;
  long cnt = r;
  while (0 < cnt) {
    bunbo *= cnt;
    bunbo % mod;
    cnt--;
  }
  if (bunbo == 0) return 1;
  long long bunsi = 1;
  cnt = 0;
  while (cnt < r) {
    bunsi *= (n - cnt);
    bunbo %= mod;
    cnt++;
  }
  return bunsi / bunbo % mod;
}
int main() {
  int n;
  cin >> n;
  vector<long> a(n);
  REP(i, n) cin >> a[i];
  sort(a.begin(), a.end());
  ll _n = a[n - 1];
  ll _r = a[0];
  ll avg = _n / 2, avg2 = (_n + 1) / 2, nowbib = (ll)INF*10;
  REP(i, n - 1) {
    if (abs(avg +_n - a[i]) < nowbib || abs(avg2 +_n- a[i]) < nowbib) {
      _r = a[i];
      nowbib = abs(avg - _n + a[i]);
    }
  }
  cout << _n << " " << _r << endl;
}