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: Accepted)

#include<cstdlib>
#include<iostream>
#include<cmath>
#include<iostream>
#define inf 1000000001LL
#define ll long long

using namespace std;

long long lx=inf,rx=-inf,uy=-inf,dy=inf;

long long _min(long long a,long long b){if(a<b)return a; return b;}
long long _max(long long a,long long b){if(a>b)return a; return b;}

struct node{ll y; ll x;}p[60];

int main(){
	int N,K,cnt;
	long long Ans=5e18;
	scanf("%d%d",&N,&K);
	int i,j,k,l,ii;
	
	for(i=1;i<=N;i++)
		scanf("%lld%lld",&p[i].x,&p[i].y);
	
	for(i=1;i<=N;i++)
		for(j=i;j<=N;j++)
			for(k=j;k<=N;k++)
				for(l=k;l<=N;l++){
					lx=_min(_min(p[i].x,p[j].x),_min(p[k].x,p[l].x));
					rx=_max(_max(p[i].x,p[j].x),_max(p[k].x,p[l].x));
					uy=_max(_max(p[k].y,p[l].y),_max(p[i].y,p[j].y));
					dy=_min(_min(p[i].y,p[j].y),_min(p[k].y,p[l].y));
					cnt=0;
					for(ii=1;ii<=N;ii++)
						if(p[ii].x>=lx&&p[ii].x<=rx&&p[ii].y>=dy&&p[ii].y<=uy)cnt++;
					//printf("%d\n",cnt);
					if(cnt>=K&&(rx-lx)*(uy-dy)<Ans)Ans=(rx-lx)*(uy-dy);
				}
	printf("%lld",Ans);
	return 0;
}