Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 4
1 4
3 3
6 2
8 1

Sample Output 1

21

One rectangle that satisfies the condition with the minimum possible area has the following vertices: (1,1), (8,1), (1,4) and (8,4).

Its area is (8-1) × (4-1) = 21.

Sample Input 2

4 2
0 0
1 1
2 2
3 3

Sample Output 2

1

Sample Input 3

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

Sample Output 3

3999999996000000001

Watch out for integer overflows.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;

int main(){
    typedef pair<int, int> P;
    int n, k;
    cin >> n >> k;
    vector<P> vertex(n);
    vector<int> vertex_x(n);
    vector<int> vertex_y(n);
    int x, y;
    for (int i = 0; i < n; i++) {
        cin >> x >> y;
        vertex[i] = {x, y};
        vertex_x[i] = x;
        vertex_y[i] = y;
    }
    sort(vertex_x.begin(), vertex_x.end());
    sort(vertex_y.begin(), vertex_y.end());

    long long ans = 4'000'000'000'000'000'000;
    long long x1, x2, y1, y2;

    for (int i = 0; i < n; i++){
        for (int j = i + 1; j < n; j++){
            for (int s = 0; s < n; s++){
                for (int t = s + 1; t < n; t++){
                    x1 = vertex_x[i];
                    x2 = vertex_x[j];
                    y1 = vertex_y[s];
                    y2 = vertex_y[t];
                    int tmp = 0;
                    for (int z = 0; z < n; z++){
                        if (x1 <= vertex[z].first && vertex[z].first <= x2 && y1 <= vertex[z].second && vertex[z].second <= y2) tmp++;
                    }
                    if (tmp >= k) ans = min((long long)ans, (long long)((x2 - x1) * (y2 - y1)));
                }
            }
        }
    }
    cout << ans << endl;
}