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: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for(long long i=0; i<n; i++)
#define REPR(i, n) for(long long i=n-1; i>=0; i--)
#define FOR(i, m, n) for(long long i=m; i<=n; i++)
#define FORR(i, m, n) for(long long i=m; i>=n; i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VSORTR(v) sort(v.rbegin(), v.rend());
#define ALL(v) (v).begin(),(v).end()
#define FIN ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

using namespace std;
using ll = long long;
using ull = unsigned long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<ll, ll>;
const ll mod = 1e9+7;
const ll inf = 1e15;
ll power(ll a,ll b){return b?power(a*a%mod,b/2)*(b%2?a:1)%mod:1;}


char a[55][55];

int main(){FIN

    ll n;cin>>n;
    vector<double> a(n); REP(i,n) cin>>a[i];
    VSORT(a);
    double mx = a[n-1];
    double hal = a[n-1]/2;
    double base=double(mod);
    double ans=a[0];
    REP(i,n-1){
        if (abs(a[i] - hal)<base){
            ans=a[i];
            base = abs(a[i]-hal);
        }
    }
    cout<<ll(a[n-1])<<' '<<ll(ans)<<endl;
    return 0;
}