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 <iostream>
#include <functional>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

int n,k;
vector<int> x,y;
long long  ans;

int main(){
    int xt[51], yt[51];
    cin >> n >> k;
    for(int i = 0; i < n; i++){
        cin >> xt[i] >> yt[i];
        x.push_back(xt[i]);
        y.push_back(yt[i]);
    }

    sort(x.begin(), x.end());
    sort(y.begin(), y.end());
    
    ans = 1LL * (x[n-1] - x[0]) * (y[n-1] - y[0]);
    for(int xi = 0; xi < n; xi++){
        for(int xj = xi + 1; xj < n; xj++){
            for(int yi = 0; yi < n; yi++){
                for(int yj = yi+1; yj < n; yj++){
                    int c=0;
                    for(int i = 0; i < n; i++){
                        if(x[xi] <= xt[i] && x[xj] >= xt[i] && y[yi] <= yt[i] && y[yj] >= yt[i])    c++;
                    }
                    if(k <= c){
                        ans = min(ans,1LL * (x[xj]-x[xi]) * (y[yj]-y[yi]));
                    }
                }
            }
        }
    }

    cout << ans << '\n';
}