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 ncr(int n, int r);
int factorial(int n);
typedef int value_type; /* ソートするキーの型 */

value_type med3(value_type x, value_type y, value_type z)
/* x, y, z の中間値を返す */
{
    if (x < y)
        if (y < z) return y; else if (z < x) return x; else return z; else
        if (z < y) return y; else if (x < z) return x; else return z;
}

void quicksort(value_type a[], int left, int right)
/* クイックソート
 * a     : ソートする配列
 * left  : ソートするデータの開始位置
 * right : ソートするデータの終了位置
 */
{
    if (left < right) {
        int i = left, j = right;
        value_type tmp, pivot = med3(a[i], a[i + (j - i) / 2], a[j]); /* (i+j)/2ではオーバーフローしてしまう */
        while (1) { /* a[] を pivot 以上と以下の集まりに分割する */
            while (a[i] < pivot) i++; /* a[i] >= pivot となる位置を検索 */
            while (pivot < a[j]) j--; /* a[j] <= pivot となる位置を検索 */
            if (i >= j) break;
            tmp = a[i]; a[i] = a[j]; a[j] = tmp; /* a[i],a[j] を交換 */
            i++; j--;
        }
        quicksort(a, left, i - 1);  /* 分割した左を再帰的にソート */
        quicksort(a, j + 1, right); /* 分割した右を再帰的にソート */
    }
}

//int factN;
int max;
int ind[2];

int main(){
    int n;
    scanf("%d", &n);
    int a[n];
    for(int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    quicksort(a, 0, n-1);
    for(int j = 1; j < n; j++){
        for(int i = 0; i < j; i++){
            int c = ncr(a[j], a[i]);
            if(max < c){
                max = c;
                ind[1] = j;
                ind[2] = i;
            }
        }
    }
    printf("%d %d\n", a[ind[1]], a[ind[2]]);
    return 0;
}



int ncr(int n, int r){
    int a;
    a = factorial(n)/(factorial(n - r) * factorial(r));
}

int factorial(int n){
    int x = n;
    while(n > 1){
        n = n-1;
        x = x * n;
    }
    return x;
}