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>
typedef long long ll;
 
using namespace std;

int main(){
	ll N, K;
	cin >> N >> K;
	vector<ll> x(N), y(N);
	for (int i = 0; i < N; i++){
		cin >> x[i] >> y[i];
	}

	set<ll> X;
	set<ll> Y;
	for (int i = 0; i < N; i++){
		X.insert(x[i]);
		Y.insert(y[i]);
	}

	ll res = 1e18*4;
	for (auto l = X.begin(); l != X.end(); l++){
		for (auto r = X.begin(); r != X.end(); r++){
			for (auto b = Y.begin(); b != Y.end(); b++){
				for (auto t = Y.begin(); t != Y.end(); t++){
					ll cnt = 0;
					for (int i = 0; i < N; i++){
						if (x[i] >= *l && x[i] <= *r){
							if (y[i] >= *b && y[i] <= *t){
								cnt++;
							}
						}
					}

					if (cnt >= K){
						ll s = (*r-*l)*(*t-*b);
						res = min(res, s);
					}
				}
			}
		}
	}

	cout << res << endl;
	return 0;
}