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 <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <memory>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

// c++11
#include <array>
#include <tuple>
#include <unordered_map>
#include <unordered_set>

#define mp make_pair
#define mt make_tuple
#define rep(i, n) for (int i = 0; i < (n); i++)

using namespace std;

using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;

const int INF = 1 << 29;
const ll LL_INF = 1LL << 60;
const double EPS = 1e-9;
const ll MOD = 1000000007;

const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
int N;
vector<ll> A;
int main() {
  cin >> N;
  A.resize(N);
  for (auto &val : A) {
    cin >> val;
  }
  sort(A.rbegin(), A.rend());
  int a = A[0];
  int result = INF;
  int b = A[1];
  for (int i = 1; i < N; i++) {
    int c = A[i];
    int res = abs(a - 2 * c);
    if (res <= result) {
        result = res;
        b = c;
    }
  }
  cout << a << " " << b << endl;
  return 0;
}