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 9.2.1) (judge verdict: Accepted)

#include<stdio.h>
void quicksort(int a[],int s,int t){
  if(s<t){
    int i=s,j=t,p=a[(s+t)/2];
    while(1){
      while(a[i]>p)i++;
      while(p>a[j])j--;
      if(i>=j)break;
      a[i]^=a[j]^=a[i]^=a[j];
      i++;j--;
    }
    quicksort(a,s,i-1);quicksort(a,j+1,t);
  }
}
int main(){
  int a[100000],n,i;
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d",&a[i]);
  quicksort(a,0,n-1);
  for(i=1;i<n;i++)if(a[i]<a[0]/2)break;
  if(i>1&&i<n)i-=a[0]/2-a[i]<a[i-1]-a[0]/2-a[0]%2?0:1;
  i-=i==n;
  printf("%d %d",a[0],a[i]);
}