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 <string.h>
#include <stdlib.h>
int i,j,n,a[300000],mid,min,o,ans,mid2;
void qt(int l,int r)
{
    int i,j,m,p;
    i=l; j=r;
    m=a[(l+r)/2];
    while (i<=j)
    {
        while (a[i]<m) i++;
        while (a[j]>m) j--;
        if (i<=j)
        {
            p=a[i];a[i]=a[j]; a[j]=p;
            i++; j--;
        }
    }
if (i<r) qt(i,r);
if (l<j) qt(l,j);
}

int main()
{
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    scanf("%d",&a[i]);
    qt(1,n);
    mid=a[n]/2;
    if (a[n]%2==1)
    {
        mid2=a[n]/2+1;
        min=1000000010;
        for (i=1;i<=n;i++)
        {
          o=a[i]-mid;
          if (o<0) o=-o;
          if (o<min) {min=o; ans=a[i];}
        }
        min=1000000010;
        for (i=1;i<=n;i++)
        {
          o=a[i]-mid2;
          if (o<0) o=-o;
          if (o<min) {min=o; ans=a[i];}
        }
        printf("%d %d",a[n],ans);
    } else
   {
     min=1000000010;
      for (i=1;i<=n;i++)
      {
         o=a[i]-mid;
         if (o<0) o=-o;
         if (o<min) {min=o; ans=a[i];}
      }
      printf("%d %d",a[n],ans);
   }
    return(0);
}