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)

//    https://youtu.be/N3oCS85HvpY

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef long long unsigned llu;
typedef pair<long long, long long> pll;
const long long inf = 2000000000000000000LL;	// 2e18
#define pi                  acos(-1.0)
#define fast                ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define shesh               "\n"
#define ff                  first
#define ss                  second
#define pb                  push_back
#define bp                  pop_back
#define pf                  push_front
#define fp                  pop_front
#define ub                  upper_bound
#define lb                  lower_bound
#define all(x)              x.begin(), x.end()
#define debug(x)            cout<<"debug "<<x<<"\n"
#define mest(a,b)           memset(a,b,sizeof(a))

int main()
{
    fast;
    llu ans=5000000000000000000;
    ll t,n,m,i,j,l,k,temp,flag,x,y,x1,x2,y1,y2;
    cin>>n>>k;
    vector<pair<ll,ll> >v;
    vector<ll>X,Y;
    for(i=0; i<n; i++)
    {
        cin>>x>>y;
        v.pb({x,y});
        X.pb(x);
        Y.pb(y);
    }
    for(j=0; j<n; j++)
    {
        for(x=0; x<n; x++)
        {
            for(y=0; y<n; y++)
            {
                for(t=0,temp=0; t<=n; t++)
                {
                    x1=min(X[j],X[x]);
                    x2=max(X[j],X[x]);
                    y1=min(Y[y],Y[t]);
                    y2=max(Y[y],Y[t]);
                    for(l=0,temp=0; l<n; l++)
                    {
                        if(v[l].ff>=x1 && v[l].ff<=x2 && v[l].ss>=y1 && v[l].ss<=y2)
                            temp++;
                    }
                    if(temp>=k)
                    {
                        if(ans>((llu)abs(x1-x2))*((llu)abs(y1-y2)))
                            ans=((llu)abs(x1-x2))*((llu)abs(y1-y2));
                    }
                }
            }
        }
    }
    cout<<ans;
    return 0;
}