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;
typedef long long ll;
typedef pair<int, int> P;
#define rep(i,n) for(int i = 0; i < (n); ++i)
int main() {
    int n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    if ( n == 2 ) {
        if ( a[0] > a[1] ) cout << a[0] << " " << a[1] << endl;
        else               cout << a[1] << " " << a[0] << endl;
        return 0;
    }
    sort(a.begin(), a.end());
    ll ai = a[n-1];
    int i = 0;
    bool odd = (ai & 1);
    int pre_d = 0;
    int pre_n = 0;
    while(1){
        if ( i == n - 1 ) break;
        if ( odd ) {
            if ( a[n-1-i] <= ai/2 + 1 ) {
                if ( a[n-1-i] == ai/2 + 1 || a[n-1-i] == ai/2 ) {
                    cout << ai << " " << a[n-1-i] << endl;
                    return 0;
                }
                else if ( pre_d < ( ai/2 + 1 ) - a[n-1-i] ) {
                    cout << ai << " " << pre_n << endl;
                    return 0;
                }
                else {
                    cout << ai << " " << a[n-1-i] << endl;
                    return 0;
                }
            }
        }
        else {
            if ( a[n-1-i] <= ai/2 ) {
                if ( a[n-1-i] == ai/2 ) {
                    cout << ai << " " << a[n-1-i] << endl;
                    return 0;
                }
                else if ( pre_d < ai/2 - a[n-1-i] ) {
                    cout << ai << " " << pre_n << endl;
                    return 0;
                }
                else {
                    cout << ai << " " << a[n-1-i] << endl;
                    return 0;
                }
            }
        }
        ++i;
        pre_d = a[n-1-i] - ai/2;
        pre_n = a[n-1-i];
    }
    return 0;
}