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.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        List<Long> numberList = new ArrayList<Long>();
        long max = -1;
        for (int i = 0; i < n; i++) {
            long num = sc.nextLong();
            numberList.add(num);
            max = Math.max(max, num);
        }
        
        long r = -1;
        long maxWaru2 = max / 2;
        for (int i = 0; i < maxWaru2; i++) {
            if (numberList.contains(Long.valueOf(maxWaru2 + i))) {
                r = maxWaru2 + i;
                break;
            } else if (numberList.contains(Long.valueOf(maxWaru2 + i))) {
                r = maxWaru2 - i;
                break;
            }
        }
        
        if (r == -1) r = 0;
        System.out.println(max + " " + r);
    }
    
    public static long comb(long n, long k) {
        if (k > n / 2) {
            k = n - k;
        }
        long a = 1;
        for (long i = n; i > n - k; i--) {
            a *= i;
            System.out.println(a);
        }
        
        long b = 1;
        for (long i = k; i >= 1; i--) {
            b *= i;
        }
        
        return a / b;
    }
}