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 PI 4*atan(1)
#define INF 1e8

typedef long long ll;

int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};

int main(){
  int N, K;
  cin >> N >> K;
  vector<ll> x(N), y(N);
  for(int i = 0; i < N; i++)cin >> x[i] >> y[i];
  sort(x.begin(), x.end());
  sort(y.begin(), y.end());
  ll ans = (x[N - 1] - x[0])*(y[N - 1] - y[0]);
  for(int i = 0; i < N; i++){
    for(int j = i + 1; j < N; j++){
      for(int k = 0; k < N; k++){
        for(int l = k + 1; l < N; l++){
          int cnt = 0;
          for(int m = 0; m < N; m++){
            if(x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l]){
              cnt++;
            }
          }
          if(cnt >= K){
            ans = min(ans, (x[j] - x[i]) * (y[l] - y[k]));
          }
        }
      }
    }
  }
  cout << ans << endl;
}