Problem:
Score : 400 points

Problem Statement

We have N points in a two-dimensional plane.

The coordinates of the i-th point (1 \leq i \leq N) are (x_i,y_i).

Let us consider a rectangle whose sides are parallel to the coordinate axes that contains K or more of the N points in its interior.

Here, points on the sides of the rectangle are considered to be in the interior.

Find the minimum possible area of such a rectangle.

Constraints

2 \leq K \leq N \leq 50

-10^9 \leq x_i,y_i \leq 10^9 (1 \leq i \leq N)

x_i≠x_j (1 \leq i<j \leq N)

y_i≠y_j (1 \leq i<j \leq N)

All input values are integers. (Added at 21:50 JST)

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1
:
x_{N} y_{N}

Output

Print the minimum possible area of a rectangle that satisfies the condition.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

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

#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for (ll(i) = (m); (i) < (n); ++(i))
#define REP(i, n) FOR(i, n, 0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll)1e15;

pll G[60];

int main()
{
    int N, K;
    cin >> N >> K;
    REP(i, N)
    {
        cin >> G[i].first >> G[i].second;
    }
    sort(G, G + N, [](pll a, pll b) { return a.first == b.first ? a.second < b.second : a.first < b.first; });

    ll min = (1e18 * 4);
    REP(f, N)
    {
        FOR(l, N, f)
        {
            if (l - f + 1 < K)
                continue;
            pll ming = pll(INF, INF), maxg = pll(-INF, -INF);
            REP(i, l - f + 1)
            {
                ming.first = std::min(ming.first, G[f + i].first);
                ming.second = std::min(ming.second, G[f + i].second);
                maxg.first = std::max(maxg.first, G[f + i].first);
                maxg.second = std::max(maxg.second, G[f + i].second);
            }
            min = std::min(min, (abs(maxg.first - ming.first) * abs(maxg.second - ming.second)));
        }
    }
    cout << min << endl;
    return 0;
}