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;

int binaryS(vector<int> v,int a){
    int right=v.size()-1;
    int left=0;
    int mid,ans;
    while(1){
        if(right-left==1){
            if((v[right]-a)<(a-v[left]))
                ans=v[right];
            else
                ans=v[left];
            break;
        }
        mid=(left+right)/2;
        if(v[mid]>a)
            right=mid;
        else if(v[mid]<a)
            left=mid;
        else{
            ans=a;
            break;
        }
    }

    return ans; 
}

int main(){
    int n;
    cin>>n;
    vector<int> v;
    for(int i = 0; i < n; ++i){
        int tmp;
        cin>>tmp;
        v.push_back(tmp);
    }

    sort(v.begin(), v.end());
    if(n==2)
    {
        cout<<v[1]<< " "<<v[0];
        cout<<endl;
        return 0;
    }


    int mx=v[n-1];
    int mm=(mx+1)/2;
    v.pop_back();

    cout<<mx<< " "<<binaryS(v,mm);
    cout<<endl;


    return 0;
    
}