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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val n = readInt()
    val a = readIntList().sorted()
    val r = a.last() / 2.0
    var prev = a[a.lastIndex - 1]
    for (i in n - 3 downTo 1) {
        if (Math.abs(a[i] - r) > Math.abs(prev - r)) {
            println("${a.last()} $prev")
            return
        } else {
            prev = a[i]
        }
    }
    println("${a.last()} ${a[0]}")
}

fun printYesOrNo(flag: Boolean) {
    println(if (flag) "YES" else "NO")
}

fun readIntList(): List<Int> = readLine()!!.split(' ').map { it.toInt() }

fun readInt(): Int = readLine()!!.toInt()

fun err(print: Any) {
    System.err.println(print)
}