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<iostream>
#include<set>
using namespace std;

typedef long long ll;

int n;
set<int> A;

int main() {
    cin >> n;
    for(int i = 0; i < n; i++) {
        int a;
        cin >> a;
        A.insert(a);
    }

    int mx = *A.rbegin();
    A.erase(mx);
    auto iter = A.lower_bound(mx/2);
    if(iter == A.begin()) {
        cout << mx << " " << *iter << endl;
        return 0;
    }
    if(iter == A.end()) {
        cout << mx << " " << *(--iter) << endl;
        return 0;
    }
    int r = *iter;
    int l = *(--iter);
    int ans = (mx - l < r - mx) ? l : r;

    cout << l << " " << r << endl;

    cout << mx << " " << ans << endl;

    return 0;
}