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: Accepted)

#include <bits/stdc++.h>

using namespace std;

constexpr int MOD = 1000000007;

using Long = long long;
template<typename T>
using V = vector<T>;
template<typename T>
using Vv = V<V<T>>;
using Vb = V<bool>;
using Vvb = V<Vb>;
using Vi = V<int>;
using Vvi = V<Vi>;
using Vl = V<Long>;
using Vvl = V<Vl>;
using Vd = V<double>;
using Vvd = V<Vd>;
using Vs = V<string>;

template <typename First, typename Second>
istream& operator>>(istream& is, pair<First, Second>& v);

template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
    for (auto& x : v) {
        is >> x;
    }
    return is;
}

template <typename First, typename Second>
istream& operator>>(istream& is, pair<First, Second>& v) {
    is >> v.first >> v.second;
    return is;
}

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
    for (const auto& x : v) {
        os << x << ' ';
    }
    return os;
}

// *** SOLUTION BEGIN ***

void solve() {
    int n;
    cin >> n;
    Vi a(n);
    cin >> a;
    sort(begin(a), end(a));
    int m = a.back();
    a.pop_back();
    int best = a.front();
    for (int x : a) {
        if (abs(2LL * x - m) < abs(2LL * best - m)) {
            best = x;
        }
    }
    cout << m << ' ' << best;
}

// *** SOLUTION END ***

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    cout.precision(10);
    cout << fixed;
    solve();
    return 0;
}