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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        long[] x = new long[n];
        long[] y = new long[n];
        long[][] p = new long[n][2];
        for (int i = 0; i < n; i++) {
            x[i] = sc.nextLong();
            y[i] = sc.nextLong();
            p[i] = new long[]{x[i], y[i]};
        }
        Arrays.sort(x);
        Arrays.sort(y);
        Arrays.sort(p, (a,b)->Long.compare(a[0], b[0]));
        int[][] a = new int[n][n];
        for (int i = 0; i < n; i++) {
            long ty = p[i][1];
            for (int j = 0; j < n; j++) {
                if(0 < i) a[i][j] += a[i-1][j];
                if(0 < j) a[i][j] += a[i][j-1];
                if(0 < i && 0 < j) a[i][j] -= a[i-1][j-1];
                if(ty == y[j]) a[i][j]++;
            }
        }
        long ans = Long.MAX_VALUE;
        for (int i = 0; i <= n-k; i++) {
            for (int j = 0; j <= n-k; j++) {
                for (int ii = k-1; ii < n-i; ii++) {
                    for (int jj = k-1; jj < n-j; jj++) {
                        int c = a[i+ii][j+jj];
                        if(0 < i) c -= a[i-1][j+jj];
                        if(0 < j) c -= a[i+ii][j-1];
                        if(0 < i && 0 < j) c += a[i-1][j-1];
                        if(c < k)continue;
                        long tmp = (x[i+ii] - x[i]) * (y[j+jj] - y[j]);
                        if(tmp < ans) ans = tmp;
                    }
                }
            }
        }
        System.out.println(ans);
        sc.close();

    }

}