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;
typedef long long lint;
typedef pair<int, int> P;

int ss[55][55];
int x[55], y[55];
P point[55];

int query(int x1, int y1, int x2, int y2){
    return ss[x2][y2] - ss[x2][y1] - ss[x1][y2] + ss[x1][y1];
}

lint solve(int x1, int y1, int x2, int y2){
    return 1LL * (x[x2-1]-x[x1]) * (y[y2-1]-y[y1]);
}

int main(){
    int n, k; cin >> n >> k;
    for(int i=0;i<n;i++){
        int xx, yy; cin >> xx >> yy;
        x[i] = xx; y[i] = yy;
        point[i] = P(xx, yy);
    }

    sort(x, x+n), sort(y, y+n);

    for(int i=0;i<n;i++){
        int posx, posy; P p = point[i];
        posx = distance(x, lower_bound(x, x+n, p.first));
        posy = distance(y, lower_bound(y, y+n, p.second));
        ss[posx+1][posy+1]++;
    }

    for(int i=0;i<=n;i++)
        for(int j=0;j<n;j++) ss[i][j+1] += ss[i][j];

    for(int i=0;i<n;i++)
        for(int j=0;j<=n;j++) ss[i+1][j] += ss[i][j];

    lint ans = 1e20;
    for(int x1=0;x1<=n;x1++){
        for(int x2=x1+1;x2<=n;x2++){
            for(int y1=0;y1<=n;y1++){
                for(int y2=y1+1;y2<=n;y2++){
                    if(k<=query(x1, y1, x2, y2)) ans = min(ans, solve(x1, y1, x2, y2));
                }
            }
        }
    }

    cout << ans << endl;
    return 0;
}