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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vii = vector<vi>;
using in6 = int64_t;
using ind = double;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int beki(int x) {
  int ret = 1;
  while(x) {
    ret *= x;
    x--;
  }
  return ret;
}

int comb(int x,int y) {
  int ret = beki(x) / beki(x - y) /beki(y);
  return ret;
}

int main() {
  int n;
  cin >> n;
  vi data(n);
  rep(i,n) cin >> data[i];
  sort(data.begin(), data.end());
  int ans = 0;
  pair<int,int> pi;
  rep(i,n) {
    rep(j,i) {
      int a = comb(data[i],data[j]);
      if(a >= ans) {
        pi.first = data[i];
        pi.second = data[j];
      }
      ans = max(ans,a);
    }
  }
  cout << pi.first << " " << pi.second << endl;
}