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<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll>pii;
int main()
{
	int num, mok;
	scanf("%d%d", &num, &mok);
	vector<pii>v;
	vector<ll>x, y;
	for (int i = 0; i < num; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		v.push_back(make_pair(za, zb));
		x.push_back(za), y.push_back(zb);
	}
	sort(v.begin(), v.end());
	ll mini = 8000000000000000000LL;
	for (int i = 0; i < num; i++)
	{
		for (int j = 0; j < num; j++)
		{
			for (int k = 0; k < num; k++)
			{
				ll lx = x[i], ly = y[j], uy = y[k];
				int cnt = 0;
				for (int l = 0; l < num; l++)
				{
					if (lx <= v[l].first&&ly <= v[l].second&&v[l].second <= uy)cnt++;
					if (cnt == mok)mini = min(mini, (v[l].first - lx)*(uy - ly));
				}
			}
		}
	}
	printf("%lld\n", mini);
}