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 <iostream>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<list>
#include<string>
#include<algorithm>
#include<cstdio>
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define mp make_pair
#define pb push_back
#define P pair<int,int>
#define ll __int64
#define __int64 long long

int k,n;
ll x[100],y[100];


int main(){	
	cin>>n>>k;
	ll ans=100000000000000000;
	rep(i,n)cin>>x[i]>>y[i];


	rep(i,n){
		rep(j,n){
		//	if(i==j)continue;
			rep(k,n){
				//if(i==k||j==k)continue;
				rep(l,n){
					//if(i==l||j==l||k==l)continue;
					int cnt=0;
					rep(z,n){
						if(x[i]<=x[z]&&x[z]<=x[j]&&y[k]<=y[z]&&y[z]<=y[l])cnt++;
					}

					if(cnt>=k){
						ans=min(ans,(x[j]-x[i])*(y[l]-y[k]));
						cout<<x[i]<<" "<<x[j]<<" "<<y[k]<<" "<<y[l]<<" "<<cnt<<endl;
					}
				}
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}