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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DAxisParallelRectangle solver = new DAxisParallelRectangle();
        solver.solve(1, in, out);
        out.close();
    }

    static class DAxisParallelRectangle {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            long[] x = new long[n];
            long[] y = new long[n];
            for (int i = 0; i < n; i++) {
                x[i] = in.nextLong();
                y[i] = in.nextLong();
            }

            long min = Long.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    for (int l = 0; l < n; l++) {
                        for (int m = 0; m < n; m++) {
                            // i :左  j :上 m :右 l :下
                            int cnt = 0;
                            for (int o = 0; o < n; o++) {
                                if (x[i] <= x[o] &&
                                        x[o] <= x[m] &&
                                        y[l] <= y[o] &&
                                        y[o] <= y[j]) {
                                    cnt++;
                                }
                            }
                            if (cnt >= k) {
                                long s = (x[m] - x[i]) * (y[j] - y[l]);
                                min = Math.min(min, s);
                            }
                        }
                    }
                }
            }
            out.println(min);

        }

    }
}