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;

using LL = long long;
using Point = pair<LL, LL>;

LL N, K;
vector<Point> P;
vector<LL> X, Y;

LL solve() {
  LL r = 1e31;
  for(auto j = 1; j < N; ++j) for(auto i = 0; i < j; ++i) for(auto l = 1; l < N; ++l) for(auto k = 0; k < l; ++k) {
    auto lx=X[i], hx=X[j];
    auto ly=Y[k], hy=Y[l];
    auto sum=0;
    for(auto p: P) if(lx<=p.first && p.first<=hx && ly<=p.second && p.second<=hy) ++sum;
    if(K<=sum) r=min(r, (hx-lx)*(hy-ly));
  }
  return r;
}

int main() {
  cin >> N >> K;
  for(auto i = 0; i < N; ++i) {
    LL x, y;
    cin >> x >> y;
    P.emplace_back(x, y);
    X.emplace_back(x);
    Y.emplace_back(y);
  }
  sort(begin(X), end(X));
  sort(begin(Y), end(Y));
  cout << solve() << endl;
}