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)

// https://beta.atcoder.jp/contests/abc075
// status:

#define SUBMIT
//#define DEBUG

#include <utility>
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <queue>
#include <numeric>
#include <limits>

using namespace std;
using ui64 = unsigned long long;
using i64 = long long;

int main() {
#ifdef SUBMIT
    auto& stream = cin;
#else
    auto program_input = std::string(R"(4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999
)");
    stringstream stream(program_input);
#endif
    int N, K;
    stream >> N >> K;

    vector<int> x(N), y(N);
    for (int i = 0; i < N; ++i) stream >> x[i] >> y[i];

    sort(x.begin(), x.end());
    sort(y.begin(), y.end());

    ui64 ans = static_cast<ui64>(x[x.size() - 1] - x[0]) * static_cast<ui64>(y[y.size() - 1] - y[0]);
    for (int xi = 0; xi < N; ++xi) {
        for (int xj = xi + 1; xj < N; ++xj) {
            for (int yi = 0; yi < N; ++yi) {
                for (int yj = yi + 1; yj < N; ++yj) {
                    int n_contains = 0;
                    for (int i = 0; i < N; ++i) {
                        if (x[xi] <= x[i] && x[i] <= x[xj] && y[yi] <= y[i] && y[i] <= y[yj]) n_contains++;
                    }

                    if (K <= n_contains) ans = min(ans, static_cast<ui64>(x[xj] - x[xi]) * static_cast<ui64>(y[yj] - y[yi]));
                }
            }
        }
    }

    cout << ans << endl;
    return 0;
}