Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define inf 1000000007

vector<pair<ll,ll>>a;
int main() {
	ll n,k,ans=4999999996000000001;
	cin>>n>>k;
	for(int i=0;i<n;i++){
		int x,y;
		cin>>x>>y;
		a.push_back(make_pair(x,y));
	}
	sort(a.begin(),a.end());
	for(int i=0;i<n-k+1;i++){
		ll minx=10000000000,miny=10000000000;
		ll maxx=-10000000000,maxy=-10000000000;
		for(int j=i;j<k+i;j++){
			maxx=max(maxx,a[j].first);
			minx=min(minx,a[j].first);
			maxy=max(maxy,a[j].second);
			miny=min(miny,a[j].second);
		}
		ans=min(ans,(maxx-minx)*(maxy-miny));
	}
	cout <<ans;
	return 0;
}