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: Time Limit Exceeded)

#include <iostream>
using namespace std;
#include <algorithm>
#include <vector>

#define INT int
#define REP(i,a,b) for(int i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

int vector_finder(std::vector<int> vec, int number) {
    vector<int>::iterator itr = find(vec.begin(), vec.end(), number);
    size_t index = std::distance( vec.begin(), itr );
    if (index != vec.size()) { // 発見できたとき
        return 1;
    }
    else { // 発見できなかったとき
        return 0;
    }
}

int main(){
    vector<int> a;
    int n;
    cin >> n;
    
    rep(i,n){
        int in;
        cin >> in;
        a.push_back(in);
    }
    sort(a.begin(), a.end());
    
    int out[2];
    out[0] = a[n-1];
    a[n-1] = 0;
    out[1]=out[0]/2;
        
    for(int i = 0; ; i++){
        if(vector_finder(a, out[1]+i)) {out[1] += i;break;}
        if(vector_finder(a, out[1]-i)) {out[1] -= i;break;}
    }
    cout << out[0] << " " << out[1] <<endl;
    return 0;
}