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>


/*
int main(void){
    int n;
    scanf("%d",&n);
    int a[n];
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    int large, small;
    long long int num = 0;
    long long int max = 0;
    for(int i=0;i<n;i++){
        for(int j=i+1;j<n;j++){
            if(a[i]<a[j]){
                num = comb(a[j],a[i]);
                if(max<num){
                    max = num;
                    large = a[j];
                    small = a[i];
                }
            }else{
                num = comb(a[i],a[j]);
                if(max<num){
                    max = num;
                    large = a[i];
                    small = a[j];
                }
            }
            
        }
    }
    printf("%d %d",large,small);
    return 0;
}
*/

int main(void){
    int n;
    scanf("%d",&n);
    int a[n];
    int max_a=0;
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        if(max_a<a[i]){
            max_a = a[i];
        }
    }
    if(max_a%2==0){
        int half = max_a/2;
        for(int i=0;i<max_a;i++){
            for(int j=0;j<n;j++){
                if(a[j]-half<=i||a[j]-half>=-1*i){
                    if(max_a!=a[j]){
                    printf("%d %d",max_a,a[j]);
                    return 0;
                    }
                }
            }
        }
    }else{
        int halfp = max_a/2 + 1;
        int halfm = max_a/2;
        for(int i=0;i<max_a;i++){
            for(int j=0;j<n;j++){
                if(max_a!=a[j]){
                if(a[j]-halfm<=i||a[j]-halfp>=-1*i){
                    printf("%d %d",max_a,a[j]);
                    return 0;
                }
                }
            }
        }       
    }
    return 0;
}