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;
using ll = long long;
const ll INF = LONG_LONG_MAX / 2;

int main() {
    ll N, K; cin >> N >> K;
    vector<pair<ll,ll>> v(N);
    for (ll i = 0; i < N; i++)
    {
        cin >> v[i].first >> v[i].second;
    }
    
    ll min_area = INF;
    for (ll i = 0; i < N; i++)
    {
        for (ll j = 0; j < N; j++)
        {
            if(v[i].first>v[j].first) continue;
            for (ll k = 0; k < N; k++)
            {
                for (ll l = 0; l < N; l++)
                {
                    if(v[k].second>v[l].second) continue;
                    ll area = (v[j].first-v[i].first) * (v[l].second-v[k].second); 
                    if(area > min_area) continue;
                    ll cnt = 0;
                    for (ll m = 0; m < N; m++)
                    {
                        if(v[i].first<=v[m].first && v[j].first>=v[m].first && v[k].second<=v[m].second && v[l].second>=v[m].second){
                          cnt++;  
                        }                        
                    }
                    if(cnt>=K) min_area = min(min_area, area);
                }
            }
        }
    }
    cout << min_area << endl;
}