Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

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


Output

3999999996000000001

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int N = sc.nextInt();
        int K = sc.nextInt();
        int[] x = new int[N];
        int[] y = new int[N];
        for (int i = 0; i < N; ++i) {
            x[i] = sc.nextInt();
            y[i] = sc.nextInt();
        }

        System.out.println(solve(x, y, K));

        sc.close();
    }

    static long solve(int[] x, int[] y, int K) {
        long result = Long.MAX_VALUE;
        for (int minX : x) {
            for (int maxX : x) {
                for (int minY : y) {
                    for (int maxY : y) {
                        int interiorCount = 0;
                        for (int i = 0; i < x.length; ++i) {
                            if (x[i] >= minX && x[i] <= maxX && y[i] >= minY && y[i] <= maxY) {
                                ++interiorCount;
                            }
                        }

                        if (interiorCount >= K) {
                            result = Math.min(result, (long) (maxX - minX) * (maxY - minY));
                        }
                    }
                }
            }
        }

        return result;
    }
}