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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    //D - Binomial Coefficients
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        long[] ai = new long[n];
        for(int i = 0; i < n; i++) {
            ai[i] = scanner.nextLong();
        }
        Arrays.sort(ai);

        BigInteger maxComb = BigInteger.ZERO;
        int maxI = 0;
        int maxJ = 0;
        for(int j = 0; j < n - 1; j++) {
            for(int i = j + 1; i < n; i++) {
                BigInteger c = comb(ai[i], ai[j]);
                if ( maxComb.compareTo(c) < 0) {
                    maxComb = c;
                    maxI = i;
                    maxJ = j;
                }
            }
        }
        System.out.printf("%d %d\n", ai[maxI], ai[maxJ]);
    }

    private static BigInteger comb(long n, long r) {
        if ( r == 0 ) {
            return BigInteger.valueOf(1);
        }

        if ( r > n / 2 ) {
            return comb(n, n - r);
        }

        BigInteger p = BigInteger.ONE;
        BigInteger q = BigInteger.ONE;
        for(long i = 1; i <= r; i++) {
            p = p.multiply(BigInteger.valueOf(n - i + 1));
            q = q.multiply(BigInteger.valueOf(i));
        }
        return p.divide(q);
    }
}