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>
#include <sys/stat.h>
using namespace std;
using ll = int64_t;

int main(int , char **)
{
#ifdef LOCAL_DEBUG
	struct stat sb;
	ifstream in("/dev/shm/input.txt");
	if (stat("/dev/shm/input.txt",&sb)==0){ cin.rdbuf(in.rdbuf()); }
#endif

	int N,K;
	cin >> N >> K;
	ll px[55]={},py[55]={};

	for(int i=0;i<N;i++){
		cin >> px[i] >> py[i];
	}

	ll ans = LONG_MAX;
	for(int i=0;i<N-1;i++){
		for(int k=i+1;k<N;k++){
			ll x=px[i],y=py[i],X=px[k],Y=py[k];
			if (X<x){swap(x,X);}
			if (Y<y){swap(y,Y);}
			ll m = (X-x)*(Y-y);
			int cnt = 0;
			for(int r=0;r<N;r++){
				if ((x<=px[r])&&(px[r]<=X)&&(y<=py[r])&&(py[r]<=Y)){
					cnt++;
				}
			}
			if (cnt < K){continue;}

			ans = min(ans,m);
		}
	}

	printf("%ld\n",ans);

	return 0;
}