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<iostream>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<math.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62);

typedef long long i64;
typedef pair<i64,i64> P;


int n, k, a[101], b[101];
i64 px[101], py[101];

int main(){
	cin >> n >> k;
	for(int i = 0; i < n; i++){
		cin >> a[i] >> b[i];
		px[i] = a[i];
		py[i] = b[i];
	}
	
	sort(px, px+n);
	sort(py, py+n);
	
	i64 ans = LINF;
	for(int u = 0; u < n; u++) for(int d = 0; d < n; d++)
		for(int l = 0; l < n; l++) for(int r = 0; r < n; r++){
			int cnt = 0;
			for(int i = 0; i < n; i++){
				if(a[i] >= px[l] && a[i] <= px[r] && b[i] >= py[u] && b[i] <= py[d]){
					cnt++;
				}
			}
			if(cnt >= k) ans = min(ans, (px[r]-px[l])*(py[d]-py[u]));
		}
	
	cout << ans << endl;
	
	return 0;
}