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) {
		Main main = new Main();
		main.run();
	}

	class P implements Comparable<P>{
		long x;
		long y;
		P(long x, long y){
			this.x=x;
			this.y=y;
		}
		@Override
		public int compareTo(P arg0) {
			return Long.compare(this.x, arg0.x);
		}
	}
	public void run() {
		Scanner sc = new Scanner(System.in);
		int N=sc.nextInt();
		int K=sc.nextInt();

		long x[] = new long[N];
		long y[] = new long[N];
		for(int i=0; i<N; i++) {
			x[i]=sc.nextLong();
			y[i]=sc.nextLong();

		}

		long ans = Long.MAX_VALUE;
		//左
		for(int i=0; i<N; i++) {
			long x1 = x[i];
			//右
			for(int j=0; j<N; j++) {
				long x2 = x[j];
				if(x1>=x2) continue;
				//下
				for(int k=0; k<N; k++) {
					long y1 = y[k];
					//上
					for(int l=0; l<N; l++) {
						long y2 = y[l];
						if(y1>=y2) continue;
						int cnt=0;
						for(int m=0; m<N; m++) {
							if(x1<=x[m] && x[m]<=x2 && y1<=y[m] && y[m] <=y2) {
								cnt++;
							}
						}
						if(cnt==K) {
							ans = Math.min(ans, (x2-x1) * (y2-y1));
						}
					}
				}
			}
		}
		System.out.println(ans);
		sc.close();
	}



}