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;
typedef long long ll;
#define RAP(i, n) for(ll i=0;i<(n);i++)
#define FOR(i, a, b) for(ll i=(a);i<(b);i++)

int main(){
    ll n;
    cin >> n;
    vector<ll> A(n);
    RAP(i, n){
        cin >> A[i];
    }
    sort(A.begin(), A.end());
    ll M = A[n-1], M2, idx;
    A.pop_back();
    auto itr = lower_bound(A.begin(), A.end(), M/2);
    idx = distance(A.begin(), itr);
    if(idx==0){
        M2 = A[0]; 
    }else if(idx==(n-1)){
        M2 = A[n-2];
    }else if((A[idx]-M/2)>(M/2-A[idx-1])){
        M2 = A[idx-1];
    }else{
        M2 = A[idx];
    }
    cout << M << " " << M2 << endl;
    return 0;
}