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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long lnt;
int n, k;	lnt ans = 4e18;
struct node {int x, y;} p[100];
bool cmp(const node &a, const node &b) {return a.x < b.x;}
int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++)	scanf("%d%d", &p[i].x, &p[i].y);
	sort(p+1, p+n+1, cmp);
	for (int i = 1; i <= n; i++)	for (int j = i+k-1; j <= n; j++) {
		vector <int> tmp;
		for (int t = i; t <= j; t++)	tmp.push_back(p[t].y);
		sort(tmp.begin(), tmp.end());
		for (int t = 0; t <= tmp.size()-k; t++)	ans = min(ans, (lnt)(p[j].x-p[i].x)*(tmp[t+k-1]-tmp[t]));
	}
	printf("%lld", ans);
	return 0;
}