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>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
ll INF=9210000000000000000;
int main(){
    int n,k;cin>>n>>k;
    vector<pair<ll,ll>>v(n);
    ll x,y;rep(i,n){cin>>x>>y;v[i]=make_pair(x,y);}
    sort(v.begin(),v.end());
    ll ans=INF;
    rep(i,n-k+1){
        ll a=0-INF,b=INF;
        for(int j=i;j<i+k;j++){a=max(a,v[j].second);b=min(b,v[j].second);}
        ans=min(ans,(v[i+k-1].first-v[i].first)*(a-b));
    }
    rep(i,n)v[i]=make_pair(v[i].second,v[i].first);
    sort(v.begin(),v.end());
    rep(i,n-k+1){
        ll a=0-INF,b=INF;
        for(int j=i;j<i+k;j++){a=max(a,v[j].second);b=min(b,v[j].second);}
        ans=min(ans,(v[i+k-1].first-v[i].first)*(a-b));
    }
    cout<<ans;
}