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 javax.xml.crypto.Data;
import java.util.*;
import java.io.*;

public class Main {
    public static int N, K;
    public static int MAX = 60;
    public static long[] x = new long[MAX], y = new long[MAX];


    public static void main(String[] args) throws IOException {

        //入力
        /*
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] str = br.readLine().split(" ");
        */

        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        K = sc.nextInt();
        for(int i = 0; i < N; i++) {
            x[i] = sc.nextLong();
            y[i] = sc.nextLong();
        }
        //出力
        solve();
    }

    public static void solve() {
        long ans = 5000000000000000000L;
        for(int i = 0; i < N; i++) {
            for(int j = 0; j < N; j++) {
                for(int k = 0; k < N; k++) {
                    for(int l = 0; l < N; l++) {
                        int count = 0;
                        for(int m = 0; m < N; m++) {
                            if(x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l]) count++;
                        }
                        if(count >= K) ans = Math.min(ans, (x[j] - x[i]) * (y[l] - y[k]));
                    }
                }
            }
        }
        System.out.println(ans);

    }

}