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>

#define irep(i,n) for(int i = 0; i < (int)(n); i++)
#define irep2(i,a,n) for(int i = (int)(a); i <= (int)(n); i++)
#define lrep(i,n) for(long long  i = 0; i < (long long)(n); i++)
#define lrep2(i,a,n) for(long long i = (long long)(a); i <= (long long)(n); i++)
#define irrep(i,n) for(int i = (int)(n-1); i > -1; i--)
#define irrep2(i,a,n) for(int i = (int)(n); i >= (int)(a); i--)

typedef long long ll;
typedef std::vector<int> v_int;
typedef std::vector<v_int > v2_int;
typedef std::vector<ll> v_ll;
typedef std::vector<v_ll > v2_ll;
typedef std::vector<std::string> v_string;
typedef std::vector<v_string > v2_string;
typedef std::vector<bool> v_bool;
typedef std::vector<v_bool > v2_bool;
typedef std::pair<ll, ll> pll;
typedef std::pair<int, int> pii;

const double PI = 3.1415926535897932;
const int INF = (int)1e9;
const ll LINF = 1e18;
const ll MOD = 1000000007;
const int dr[4] = {1, 0, -1, 0};
const int dc[4] = {0, 1, 0, -1};

using namespace std; 

template<typename T, typename U> inline
ostream& operator<<(ostream &o, pair<T,U> &p)
{
    return o << "{" << p.first << "," << p.second << "}";
}

template<typename T> inline
ostream& operator<<(ostream &os, vector<T> &v)
{
    int len = v.size();
    for(int i = 0; i < len; i++) { 
        os << v[i];
        if(i != len-1) { os << ","; } 
    } 
    return os;
}

int main(void)
{
    int n;
    cin >> n;
    
    v_int a(n);

    irep(i, n) {
        cin >> a[i];
    }

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

    int max_a = -1;
    int b = a[n-1]/2;

    if(a[n-1] % 2 == 0) {
        irep(i, n-1) {
            if(b == max_a) { break; }
            max_a = max(max_a, a[i]);
        }
    } else {
        irep(i, n-1) {
            if(b == max_a || b+1 == max_a) { break; }
            max_a = max(max_a, a[i]);
        }
    }
    cout << a[n-1] << " " << max_a << endl;
    return 0;
}

/*atcoder*/