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)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h" // define macro "/D__MAI"

using namespace std;
typedef long long int ll;

#define xprintf(fmt,...) fprintf(stderr,fmt,__VA_ARGS__)
#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:v){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(auto cnt=0ll;cnt<(l);++cnt)
#define iterate(cnt,b,e) for(auto cnt=(b);cnt!=(e);++cnt)
#define MD 1000000007ll
#define PI 3.1415926535897932384626433832795
template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o; }
template<typename iterator> inline size_t argmin(iterator begin, iterator end) {
    return distance(begin, min_element(begin, end));
}
template<typename iterator> inline size_t argmax(iterator begin, iterator end) {
    return distance(begin, max_element(begin, end));
}
template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }

mt19937 randdev(8901016);
inline ll rand_range(ll l, ll h) {
    return uniform_int_distribution<ll>(l, h)(randdev);
}

#ifdef __MAI
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
#ifdef __VSCC
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0;
            T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc&&cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var*sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) {
            input_integer<int>(var);
            return *this;
        }
        inline MaiScanner& operator>>(long long& var) {
            input_integer<long long>(var);
            return *this;
        }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiblechar(cc); cc = getchar_unlocked());
            for (; isvisiblechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
        }
        template<typename IT> void in(IT begin, IT end) {
            for (auto it = begin; it != end; ++it) *this >> *it;
        }
    };
}
MaiScanner scanner;






int m, n, kei;

map<ll, ll> a1;

int main() {

    scanner >> n;

    priority_queue<ll> pq;
    pq.push(0);
    pq.push(0);

    repeat(i, n) {
        ll a;
        scanner >> a;
        a1[a] += 1;
        if (a1[a] == 2) {
            pq.push(a);
            a1[a] = 0;
        }
    }
    

    ll l1 = pq.top();pq.pop();
    ll l2 = pq.top();

    cout << l1*l2 << endl;
    

    return 0;
}