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 CPP):

Solution:
#include<stdio.h>
#include<climits>

int main(){
	int N;
	int K;
	scanf("%d",&N);
	scanf("%d",&K);
	long long X[50];
	long long Y[50];
	for(int i=0;i<N;i++){
		scanf("%lld", &X[i]);
		scanf("%lld", &Y[i]);
	}

	long long min=9223372036854775807;
	for(int xi=0;xi<N;xi++){
		long long xs = X[xi];
		for(int xj=0;xj<N;xj++){
			long long xe = X[xj];
			for(int yi=0;yi<N;yi++){
				long long ys = Y[yi];
				for(int yj=0;yj<N;yj++){
					long long ye = Y[yj];
					if(xs < xe && ys < ye){
						int cnt=0;
						for(int i=0;i<N;i++){
							if(xs<=X[i] && X[i]<=xe && ys<=Y[i] && Y[i]<=ye){
								cnt+=1;
							}
						}
						if(cnt>=K && min>(xe-xs)*(ye-ys)){
							min = (xe-xs)*(ye-ys);
						}
					}
				}
			}
		}
	}
	printf("%lld\n", min);
}