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: Accepted)

#include <stdio.h>
#include <string.h>

long long int temp[100000];
void MergeSort(long long int x[ ],long long int left,long long int right)
{
    long long int mid, i, j, k;

    if (left >= right)              /* 配列の要素がひとつなら */
        return;                     /* 何もしないで戻る */

                                    /* ここでは分割しているだけ */
    mid = (left + right) / 2;       /* 中央の値より */
    MergeSort(x, left, mid);        /* 左を再帰呼び出し */
    MergeSort(x, mid + 1, right);   /* 右を再帰呼び出し */

      /* x[left] から x[mid] を作業領域にコピー */
    for (i = left; i <= mid; i++)
        temp[i] = x[i];

      /* x[mid + 1] から x[right] は逆順にコピー */
    for (i = mid + 1, j = right; i <= right; i++, j--)
        temp[i] = x[j];

    i = left;         /* i とj は作業領域のデーターを */
    j = right;        /* k は配列の要素を指している */

    for (k = left; k <= right; k++)    /* 小さい方から配列に戻す */
        if (temp[i] <= temp[j])        /* ここでソートされる */
            x[k] = temp[i++];
        else
            x[k] = temp[j--];
}


int main (void){
int N;
long long int max;
long long int ans;
long long int A[100000];
scanf("%d",&N);
for (int i = 0; i < N; i++) {
        scanf("%lld",&A[i]);
}
MergeSort(A, 0, N-1);
max=A[N-1]/2;
if (N==2) {
  ans=A[0];
}else{
for (int i = 1; i < N-1; i++) {
if (max<=A[i]) {
  if (A[i]-max<=max-A[i-1]) {
    ans=A[i];
  }else{
    ans=A[i-1];
  }
  break;
}
}
}
printf("%lld %lld\n",A[N-1],ans);
        return 0;
}