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 <iostream>
#include <stdio.h>
#include <algorithm>
#include <string>
#include <cmath>
#include <vector>
#include <functional>
#include <stack>
#include <utility>
#include <map>
#include <queue>
#include <set>
#include <array>
#include <climits>
#include <cassert>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

const ll MOD = 1000000000 + 7;
const ll INF = LLONG_MAX;
const ll MAX_V = ll(1e5) + 5;
const double EPS = 10e-8;
const double PI = 3.14159265359;

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

int main() {
    int n, a[10005];
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    int amax = a[n-1], dmin = amax+1, half;
    for (int i = 0; i < n; i++) {
        if (amax % 2 == 0 && abs(a[i] - amax / 2) < dmin) {
            dmin = abs(a[i] - amax / 2);
            half = a[i];
        } else if (amax % 2 == 1 && (abs(a[i] - amax / 2) < dmin ||
                                 abs(a[i] - 1 - amax / 2) < dmin)) {
            dmin = min(abs(a[i] - amax / 2), abs(a[i] - 1 - amax / 2));
            half = a[i];
        }
    }
    cout << amax << " " << half << endl;
    return 0;
}