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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }

    static class Task {
        int N, K;
        List<long[]> P = new ArrayList<>();
        long ans = Long.MAX_VALUE;

        long calcS(List<long[]> points) {
            long minX = Long.MAX_VALUE;
            long minY = Long.MAX_VALUE;
            long maxX = Long.MIN_VALUE;
            long maxY = Long.MIN_VALUE;
            for (long[] p : points) {
                long x = p[0];
                long y = p[1];
                minX = Math.min(x, minX);
                maxX = Math.max(x, maxX);
                minY = Math.min(y, minY);
                maxY = Math.max(y, maxY);
            }

            int cnt = 0;
            for (long[] p : P) {
                if (minX <= p[0] && p[0] <= maxX && minY <= p[1] && p[1] <= maxY) {
                    cnt++;
                }
            }
            if (cnt < K) {
                return Long.MAX_VALUE;
            } else if (minX == maxX && minY == maxY) {
                return 1;
            } else if (minX == maxX) {
                return maxY - minY;
            } else if (minY == maxY) {
                return maxX - minX;
            } else {
                return (maxX - minX)*(maxY - minY);
            }
        }

        void search(List<long[]> points) {
            if (points.size() == 4) {
                ans = Math.min(ans, calcS(points));
            } else {
                for (long[] p : P) {
                    List<long[]> nlist = new ArrayList<>(points);
                    nlist.add(p);
                    search(nlist);
                }
            }
        }

        void solve(InputReader in, PrintWriter out) {
            N = in.nextInt();
            K = in.nextInt();

            for (int i = 0; i < N; ++i) {
                P.add(new long[] {in.nextLong(), in.nextLong()});
            }

            search(new ArrayList<>());

            out.println(ans);
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int size) {
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        long[] nextLongArray(int size) {
            long[] arr = new long[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }
    }
}