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: Accepted)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

int main() {
  int N, K;
  cin >> N >> K;
  vector<P> v(N);
  for (int i = 0; i < N; i++) {
    ll x, y;
    cin >> x >> y;
    v[i] = P(x, y);
  }
  sort(v.begin(), v.end(), [](P a, P b) {
    return a.first < b.first;
  });

  ll ans = -1;
  for (int l = 0; l < N-K+1; l++) {
    for (int r = l+K-1; r < N; r++) {
      vector<P> v2(r-l+1);
      copy(v.begin()+l, v.begin()+r+1, v2.begin());
      sort(v2.begin(), v2.end(), [](P a, P b) {
        return a.second < b.second;
      });
      ll dx = v[r].first - v[l].first;
      for (int i = 0; i < v2.size()-K+1; i++) {
        ll dy = v2[K-1+i].second - v2[i].second;
        ll area = dx * dy;
        ans = ans < 0 ? area : min(ans, area);
      }
    }
  }
  cout << ans << endl;
}