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;

int main()
{
    int N, K; cin >> N >> K;
    vector<pair<int, int>> xy(N); for (auto& p : xy) cin >> p.first >> p.second;
    sort(xy.begin(), xy.end());

    ll ans = LLONG_MAX;
    for (int i = 0; i < N-1; i++) {
        for (int j = i + K - 1; j < N; j++) {
            ll dx = xy[j].first - xy[i].first;
            vector<int> y;
            for (int m = i; m <= j; m++) {
                y.push_back(xy[m].second);
            }
            sort(y.begin(), y.end());
            for (int m = 0; m + K - 1 < y.size(); m++) {
                int n = m + K - 1;
                ll a = dx * (y[n] -y[m]);
                if (a < ans) ans = a;
            } 
        }
    }
    cout << ans << endl;
}