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>
using namespace std;

#define REP(i,n) for(int i=0;i<(int)(n);i++)
#define ALL(x) (x).begin(), (x).end()

typedef int64_t ll;
typedef long double ld;

const ll INF = 4e18;
const ld EPS = 1e-8;

int main(){
  int N, K;
  cin >> N >> K;
  vector<int> x(N), y(N);
  REP(i, N) cin >> x[i] >> y[i];

  ll ret = INF;
  for(int xmin : x) for (int xmax : x) {
    for (int ymin : y) for (int ymax : y) {
      if(xmin > xmax || ymin > ymax) continue;
      int cnt = 0;
      REP(i, N) if (x[i] >= xmin && x[i] <= xmax && y[i] >= ymin && y[i] <= ymax) ++cnt;
      if(cnt >= K) {
        ret = min(ret, (ll)(xmax - xmin) * (ll)(ymax - ymin));
      }
    }
  }

  cout << ret << endl;
  return 0;
}