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: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int compare(const void *a,const void *b){
    return *(int*)a-*(int*)b;
}
/*
int comb(int n, int r){
  if(r==0 || r==n)return (1);
  else if(r==1)return(n);
  return(comb(n-1,r-1)+comb(n-1,r));
}
*/
int main(){
  int n;
  int a[100001];
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
  }
  qsort(a,n,sizeof(a[0]),compare);
  float nmax=a[n-1];
  float med = nmax/2.000;
  int rmax=1111111111;
  double tempbest=1111111111;
  for(int i=0;i<n-1;i++){
    int temp=abs(med-a[i]);
    if(temp<tempbest){
      rmax=a[i];
      tempbest=temp;
    }
  }
  printf("%d %d\n",(int)nmax,rmax);
  return 0;
}