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>
#define N 55
#define X first
#define Y second
using namespace std;

typedef long long ll;
ll n, m, mn = 5e18;
pair<ll, ll> a[N];
vector<ll> v;
int main()
{
	ll i, j, k;
	cin >> n >> m;
	for (i = 0; i < n; i++) scanf("%lld %lld", &a[i].X, &a[i].Y);
	sort(a, a + n);
	for (i = 0; i < n; i++) {
		for (j = i; j < n; j++) {
			v.push_back(a[j].Y);
			sort(v.begin(), v.end());
			if (v.size() < m) continue;
			for (k = m - 1; k < v.size(); k++)
				mn = min(mn, (a[j].X - a[i].X) * (v[k] - v[k - m + 1]));
		}
		v.clear();
	}
    cout << mn << endl;
    return 0;
}