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 {
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			int n = sc.nextInt();
			int k = sc.nextInt();

			Integer [][] data = new Integer[n][2];
			for(int i = 0; i < n; i++){
				data[i][0] = sc.nextInt();
				data[i][1] = sc.nextInt();
			}
			Arrays.sort(data,new Comparator<Integer[]>() {
				@Override
				public int compare(Integer[] o1, Integer[] o2) {
					if(o1[0] < o2[0]) return -1;
					if(o1[0] > o2[0]) return 1;
					if(o1[1] < o2[1]) return -1;
					if(o1[1] > o2[1]) return 1;
					return 0;
				}
			});
			
			
			long min = Long.MAX_VALUE;
			for(int i = 0; i < n; i++){
				for(int j = 0; j < n; j++){
					for(int ii = 0; ii < n; ii++){
						for(int jj = 0; jj < n; jj++){
							
							int count = 0;
							for(int kk = 0; kk < n; kk++){
//								debug(data[kk]);
								if(data[i][0] <= data[kk][0] && data[kk][0] <= data[ii][0]
										&& data[j][1] <= data[kk][1] && data[kk][1] <= data[jj][1]){
									count++;
								}
							}
							if(count >= k){
								min = Math.min(min, ((long)data[ii][0] - data[i][0]) * ((long)data[jj][1] - data[j][1]) );
							}
						}
					}
				}
			}
			System.out.println(min);
		}
	}



	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

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

}