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 rep(i, n) for (ll i = 0; i < (n); i++)
#define all(x) (x).begin(), (x).end()
using ll = long long;
using namespace std;
template <typename T> using vec = std::vector<T>;

int main() {
    int N, K;
    cin >> N >> K;

    vec<ll> x(N), y(N),xs(N),ys(N);
    rep(i, N) cin >> x[i] >> y[i];
    xs = x;
    ys = y;
    sort(all(xs));
    sort(all(ys));

    ll ans = (1LL << 62);
    rep(i, N - 1) for (int j = i + 1; j < N; ++j)
        rep(k, N - 1) for (int l = k + 1; l < N; ++l) {
        ll xmin = xs[i], xmax = xs[j], ymin = ys[k], ymax = ys[l];
        
        int cnt = 0;
        rep(p,N){
            if(xmin <= x[p] && x[p] <= xmax && ymin <= y[p] && y[p] <= ymax) ++cnt;
        }
        if(cnt >= K) ans = min(ans,1LL*(xmax-xmin)*(ymax-ymin));
    }

    cout << ans << endl;
}