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: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int fun(int n)
{
    if(n>0)
        return n;
    else
        return -n;
}
int main()
{
    long long int i,max,min,pi,k,s1,s2,n,a[100005],num;
    while(scanf("%lld",&n)==1)
    {
        pi=0,s1=0,s2=0;
        for(i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            if(a[i]==0)
                pi=1;
            if(i==0)
                max=a[i];
            if(i!=0)
                max=max>a[i]?max:a[i];
        }
        //printf("%lld ",max);
        if(max%2==0)
        {
            s1=max/2;
            s2=s1+1;
        }
        else
            s1=(max+1)/2;
        //printf("%lld \n",s1);
        for(i=0;i<n;i++)
        {
            if(s2==0)
            {
                if(i==0)
                    min=fun(a[i]-s1),k=a[i];
                else
                {
                    if(min>fun(a[i]-s1))
                    {
                        min=fun(a[i]-s1);
                        k=a[i];
                        //printf("%lld\n",k);
                    }
                }
            }
            else
            {
                if(i==0)
                {
                    if(a[i]<=s1)
                        min=s1-a[i],k=a[i];
                    else
                        min=a[i]-s2,k=a[i];
                }
                else
                {
                    if(a[i]<=s1)
                        num=s1-a[i];
                    if(a[i]>s1)
                        num=a[i]-s2;
                    if(num<=min)
                    {

                        if(min==num)
                            k=k>a[i]?a[i]:k;
                        else
                            k=a[i];
                        min=num;
                    }
                }
            }
        }
        if(pi)
            printf("%lld %lld\n",max,0);
        else
            printf("%lld %lld\n",max,k);
    }
    return 0;
}