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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;
use std::cmp::{min, max};
use std::mem::swap;
use std::collections::HashMap;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin_lock = stdin.lock();
    let s = stdin_lock
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok().unwrap()
}

static DX: &'static [i32] = &[0, 0, 1, -1];
static DY: &'static [i32] = &[1, -1, 0, 0];

fn main() {
    let n: usize = read();

    let mut a: Vec<u32> = (0..n).map(|_| read()).collect();

    a.sort();

    let m = a[n - 1];

    let mut l = a[0];
    let mut idx = 0;

    for i in 0..n - 1 {
        let t = min(a[i], m - a[i]);

        if l < t {
            l = t;
            idx = i;
        }
    }

    println!("{} {}", m, a[idx]);
}

fn pow_mod(mut n: u64, mut exp: u64, m: u64) -> u64 {
    let mut t = 1;

    n %= m;

    while exp > 1 {
        if exp & 1 == 1 {
            t *= n;
            t %= m;
        }

        n *= n;
        n %= m;
        exp >>= 1;
    }

    (n * t) % m
}