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: Runtime Error)

#include <iostream>
#include <fstream>
#include <iomanip>
#include <math.h>
#include <limits.h>
#include <algorithm>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <string>
#include <string.h>
#include <sstream>
#include <ctime>

using namespace std;

#define eps 1e-12
#define pi 3.14159265358979323846
#define pb push_back
#define mp make_pair
#define st first
#define nd second
#define bgn begin
#define ll long long
#define ld long double
#define ull unsigned long long
#define ii pair<ll,ll>





const int N=50;
ll n,k,x[N],y[N],res,x2,x3,sz,t;
vector<ll>v;






bool cmp(int i,int j)
{
	return y[i]<y[j];
}








void solve()
{
	cin>>n>>k;
	for(int i=1;i<=n;i++)cin>>x[i]>>y[i];
	res=-1;
	for(int a=1;a<=n;a++)
	{
		for(int b=1;b<=n;b++)
		{
			x2=min(x[a],x[b]);
			x3=max(x[a],x[b]);
			v.clear();
			for(int i=1;i<=n;i++)
			{
				if(x[i]>=x2&&x[i]<=x3)v.pb(i);
			}
			sz=v.size();
			if(sz<k)continue;
			sort(v.begin(),v.end(),cmp);
			for(int i=k-1,j=0;i<sz;i++,j++)
			{
				t=(x3-x2)*(y[v[i]]-y[v[j]]);
				if(res==-1||res>t)res=t;
			}
		}
	}
	cout<<res<<"\n";
}

int main()
{
	std::ios::sync_with_stdio(0);
	cin.tie(0);
#ifdef localProject
	freopen("in.txt","r",stdin);
#endif
	solve();
	return 0;
}