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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std ;

int64_t comb(int x,int y){
  int64_t up=1,down=1 ;
  for(int i=0;i<y;i++){
    up*=(x-i) ;
    down*=(y-i) ;
  }
  return up/down ;
}

int main()
{
  int N ;
  cin>>N ;
  vector<int>a(N) ;
  for(int i=0;i<N;i++) cin>>a.at(i) ;
  sort(a.begin(),a.end()) ;
  reverse(a.begin(),a.end()) ;
 
  if(N==2){
    cout<<a.at(0)<<" "<<a.at(1)<<endl ;
    return 0 ;
  }
   
  int64_t MAX=0,center ;
  int n,m ;
  for(int i=0;i<N-1;i++){
    int max=a.at(i) ;
    if(a.at(i)%2==0){  //偶数の場合
      center=a.at(i)/2 ;
      for(int j=i+1;j<N;j++){
        if(a.at(j)>center) max=a.at(j) ;
        else if(a.at(j)=center){
          if(comb(a.at(i),center)>MAX){
            MAX=comb(a.at(i),center) ;
            n=a.at(i) ;
            m=center ;
          }
          break ;
        }else{
          max=(max-center>center-a.at(j)?a.at(j):max) ;
          if(comb(a.at(i),max)>MAX){
            MAX=comb(a.at(i),max) ;
            n=a.at(i) ;
            m=max ;
          }
          break ;
        }
      }
    }else{/*奇数の場合*/
      center=(a.at(i)+1)/2 ;
      for(int j=i+1;j<N;j++){
        if(a.at(j)>center) max=a.at(j) ;
        else if(a.at(j)==center||a.at(j)==center-1){
          if(comb(a.at(i),a.at(j))>MAX){
            MAX=comb(a.at(i),a.at(j)) ;
            n=a.at(i) ;
            m=a.at(j) ;
          }
          break ;
        }else{
          max=(max-center>center-1-a.at(j)?a.at(j):max) ;
          if(comb(a.at(i),max)>MAX){
            MAX=comb(a.at(i),max) ;
            n=a.at(i) ;
            m=max ;
          }
          break ;
        }
      }
    }
  } 
  cout<<n<<" "<<m<<endl ;
  
  return 0 ;
}