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 <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = (ll)1e9 + 7;

int main()
{
  int N, K; scanf("%d %d", &N, &K);
  vector<ll> X(N), Y(N);
  vector<pair<ll, ll>> P(N);
  for (int i = 0; i < N; ++i) {
    scanf("%lld %lld", &X[i], &Y[i]);
    P[i] = make_pair(X[i], Y[i]);
  }
  sort(X.begin(), X.end());
  sort(Y.begin(), Y.end());
  ll ans = numeric_limits<ll>::max();
  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) {
          ll a = X[i], b = X[j], c = Y[k], d = Y[l];
          int cnt = count_if(P.begin(), P.end(),
                             [a, b, c, d](pair<ll, ll> p) {
                               return a <= p.first and p.first <= b and
                               c <= p.second and p.second <= d;
                             });
          if (cnt >= K) {
            ans = min(ans, abs(X[i] - X[j]) * abs(Y[k] - Y[l]));
          }
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}