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 ll;

typedef pair<ll, ll> p;

int main() {
	cin.tie(0);
    ios::sync_with_stdio(false);

    int N, K;
    cin >> N >> K;

    vector<p> xy(N);

    for(int i = 0; i < N; i++) {
        ll x, y;
        cin >> x >> y;
        xy[i] = make_pair(x, y);
    }

    sort(xy.begin(), xy.end());

    ll ans = 4000000000000000005;

    for(int i = 0; i < N; i++) {
        for(int j = 0; j < N; j++) {
            ll y1 = xy[i].second;
            ll x1 = xy[j].first;
            for(int k = 0; k < N; k++) {
                ll y2 = xy[k].second;
                if(y1 > y2) continue;
                ll x2;
                int c = 0;
                for(int l = j; l < N; l++) {
                    if(y1 <= xy[l].second && xy[l].second <= y2) {
                        c++;
                        if(c == K) {
                            x2 = xy[l].first;
                            break;
                        }
                    }
                }
                if(c < K) continue;
                ans = min(ans, (x2 - x1) * (y2 - y1));
            }
        }
    }

    cout << ans << endl;
}