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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int cmp(const void *a, const void *b) {
  return *(int*)a > *(int*)b;
}

int main(void) {
  int n;
  int max = -1, maxi = -1, half = -1, halfi = -1;
  int *a;
  scanf("%d", &n);
  a = (int*)calloc(n, sizeof(int));
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }

  for (int i = 0; i < n; i++) {
    if (max < a[i]) {
      maxi = i;
      max = a[i];
    }
  }
  half = max;
  for (int i = 0; i < n; i++) {
    if (abs((double)max/2 - half) >= abs((double)max/2 - a[i]) && maxi != i) {
      halfi = i;
      half = a[i];
    }
  }
  printf("%d %d\n", max, half);

  return 0;
}