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 <algorithm>
#include <vector>
typedef long long ll;
#define PII pair<ll, ll>
#define F first
#define S second
using namespace std;
PII a[100];
ll n, k, ans=9e20;

int main() {
	int i, j, l;
	cin>>n>>k;
	for(i=0; i<n; i++) cin>>a[i].F>>a[i].S;
	sort(a, a+n);
	for(i=0; i<=n-k; i++) {
		vector<ll> v;
		for(j=0; j<k-1; j++) v.push_back(a[i+j].S);
		for(j=k-1; i+j<n; j++) {
			v.push_back(a[i+j].S);
			sort(v.begin(), v.end());
			for(l=0; l+k<=v.size(); l++)
				ans = min(ans, (v[l+k-1]-v[l]) * (a[i+j].F-a[i].F));
		}
	}
	cout<<ans;
	return 0;
}