Problem:
Score : 300 points

Problem Statement

We have N sticks with negligible thickness.
The length of the i-th stick is A_i.

Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides.
Find the maximum possible area of the rectangle.

Constraints

4 \leq N \leq 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible area of the rectangle.
If no rectangle can be formed, print 0.

Sample Input 1

6
3 1 2 4 2 1

Sample Output 1

2

1 \times 2 rectangle can be formed.

Sample Input 2

4
1 2 3 4

Sample Output 2

0

No rectangle can be formed.

Sample Input 3

10
3 3 3 3 4 4 4 5 5 5

Sample Output 3

20

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <algorithm>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define all(s) s.begin(), s.end()
using namespace std;
typedef long long int ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;
const int DIV = 1e9+7;
 
int main()
{
    int n;
    vi a;
 
    cin >> n;
    a.resize(n);
    rep(i,n) cin >> a[i];
 
    sort(all(a), greater<int>());
 
    int e[3];
    int m = 0;
    rrep(i,n-1){
        if(i == 0) continue;
        if(a[i] == a[i-1]){
            e[++m] = a[i];
            i++;
        }
        if(m == 2) break;
    }
    ll ans = (ll)e[1] * (ll)e[2];
    if(m < 2) puts("0");
    else cout << ans << endl;
}