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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

const int MOD = 1000000007;
const int INF = 1e9;

const int max_fac = 5 * 1e5;
long long fac[max_fac], finv[max_fac], inv[max_fac];
void ComInit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < max_fac; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

long long Comb(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main(){
  int n;
  const int max_n = 1e5;
  vector<pair<int, int>> a;

  ComInit();

  cin >> n;
  for(int i = 0; i < n; i++){
    int x;
    cin >> x;
    a.push_back(make_pair(x, i));
  }

  sort(a.begin(), a.end());

  int ans = 0;

  int i = n-1;
  int l = 0;
  int r = i - 1;
  int mid = 0;
  int trgt = a[i].first / 2;
  while(r - l > 1){
    int mid = (l + r) / 2;
    if(a[mid].first <= trgt) l = mid;
    else r = mid;
  }
  ans = a[l].first;
  if(trgt - a[l].first > a[l+1].first - trgt) ans = a[l + 1].first;

  cout << a[n-1].first << " " << ans << endl;
  return 0;
}