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 <bits/stdc++.h>
using namespace std;
#define FOR(i, n) for(int i = 0; i < (n); i++)
#define MEM(a, x) memset(a, x, sizeof(a))
#define ALL(a) a.begin(), a.end()
#define UNIQUE(a) a.erase(unique(ALL(a)), a.end())
typedef long long ll;
 
int n, k;
 
int main(int argc, char const *argv[]) {
	ios_base::sync_with_stdio(false);
	cin >> n >> k;
	vector<pair<ll, ll> > v(n);
	FOR(i, n) cin >> v[i].first >> v[i].second;
	sort(ALL(v));
	ll ans = 5e18;
	FOR(i, n) FOR(j, i-k+2) {
		ll xlen = v[i].first-v[j].first;
		vector<ll> y;
		for (int p = j; p <= i; p++) y.push_back(v[p].second);
		sort(ALL(y));
		ll ylen = 5e18;
		FOR(p, y.size()-k+1) {
			ylen = min(ylen, y[p+k-1]-y[p]);
		}
		ans = min(ans, xlen*ylen);
	}
	cout << ans << endl;
	return 0;
}