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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

class Main {

    public static void main(String[] args) {
        new Main().compute();
    }

    void compute() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int K = sc.nextInt();
        Point[] points = new Point[N];
        for (int i = 0; i < N; i++) {
            long x = sc.nextLong();
            long y = sc.nextLong();
            points[i] = new Point(x, y);
        }
        Point[] xSortedPoints = points.clone();
        Arrays.sort(xSortedPoints, (o1, o2) -> (int) (Math.signum(o1.x - o2.x)));
        Point[] ySortedPoints = points.clone();
        Arrays.sort(ySortedPoints, (o1, o2) -> (int) (Math.signum(o1.y - o2.y)));
        long minArea = (long) 2000000000 * 2000000000 + 1;
        for (int i = 0; i < N - K + 1; i++) {
            long xmax = -1000000001, ymax = -1000000001, xmin = 1000000001, ymin = 1000000001;
            for (int j = i; j < i + K; j++) {
                xmax = xmax > xSortedPoints[j].x ? xmax : xSortedPoints[j].x;
                ymax = ymax > xSortedPoints[j].y ? ymax : xSortedPoints[j].y;
                xmin = xmin < xSortedPoints[j].x ? xmin : xSortedPoints[j].x;
                ymin = ymin < xSortedPoints[j].y ? ymin : xSortedPoints[j].y;
            }
            long currentArea = (xmax - xmin) * (ymax - ymin);
            minArea = minArea < currentArea ? minArea : currentArea;
        }
        for (int i = 0; i < N - K + 1; i++) {
            long xmax = -1000000001, ymax = -1000000001, xmin = 1000000001, ymin = 1000000001;
            for (int j = i; j < i + K; j++) {
                xmax = xmax > ySortedPoints[j].x ? xmax : ySortedPoints[j].x;
                ymax = ymax > ySortedPoints[j].y ? ymax : ySortedPoints[j].y;
                xmin = xmin < ySortedPoints[j].x ? xmin : ySortedPoints[j].x;
                ymin = ymin < ySortedPoints[j].y ? ymin : ySortedPoints[j].y;
            }
            long currentArea = (xmax - xmin) * (ymax - ymin);
            minArea = minArea < currentArea ? minArea : currentArea;
        }
        System.out.println(minArea);
    }

    class Point {

        long x;
        long y;

        Point(long x, long y) {
            this.x = x;
            this.y = y;
        }
    }

}