Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define MOD 1000000007
#define INF 1001001001
#define LINF 4001001001001001001
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define bit(n) (1LL<<(n))
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef pair<ll, ll> LLP;

int main() {

  int N, K;
  cin >> N >> K;
  vector<LLP> p(N);
  rep(i,N) cin >> p[i].first >> p[i].second;

  sort(p.begin(), p.end());

  ll ans = LINF;

  for (int i = 0; i < N; i++) {
    for (int j = i + K - 1; j < N; j++) {
      vector<ll> y;
      for (int k = i; k <= j; k++) y.push_back(p[k].second);
      sort(y.begin(), y.end());
      for (int k = 0; k + K - 1 < y.size(); k++) {
        ll area = (p[j].first - p[i].first) * (y[k + K - 1] - y[k]);
        ans = min(ans, area);
      }
    }
  }

  cout << ans << endl;

  return 0;
}