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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define mod 1000000007
typedef long long int ll;
int u(const void *a, const void *b){
    return *(ll*)a-*(ll*)b;
}
int d(const void *a, const void *b){
    return *(ll*)b-*(ll*)a;
}
int z(int a){
  if(a<1)a*=-1;
  return a;
}
int min(int a,int b){
  if(a>b)return b;
  return a;
}
int max(int a,int b){
  if(a>b)return a;
  return b;
}
int gcd(int a,int b){
  if(!b)return a;
  return gcd(b,a%b);
}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
int n;
int f(int b,int c){
  return c*((n-1)/b+1);
}
int main(void){
  int n;
  scanf("%d",&n);
  ll a[n],m=0,ans=0;
  FOR(n){
    scanf("%lld",&a[i]);
    m=max(m,a[i]);
  }
  FOR(n)if(ans<=min(a[i],m-a[i]))ans=a[i];
  printf("%lld %lld\n",m,ans);
  return 0;
}