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: Wrong Answer)

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 105;
ll x[maxn], y[maxn];
struct Point
{
	int x, y;
};
Point p[maxn];
vector<ll>vx, vy;
int N, K;

int cmp(Point p1, Point p2)
{
	if (p1.x == p2.x)
		return p1.y < p2.y;
	else
		return p1.x < p2.x;
}

int main()
{
	cin >> N >> K;
	for (int i = 0; i<N; i++) {
		cin >> p[i].x >> p[i].y;
	}
	sort(p, p + N, cmp);
	ll ans = 1ll << 62;
	for (int i = 0; i < N; i++)
	{
		for (int j = i + 1; j < N; j++) {
			int t = 0;
			ll h = -ans;
			ll l = ans;
			for (int k = 0; k < N; k++) {
				if (p[k].x >= p[i].x&&p[k].x <= p[j].x) t++;
				h = max((ll)p[k].y, h);
				l = min((ll)p[k].y, l);
				if (t >= K) {
					ans = min(ans, (ll)(p[j].x - p[i].x)*(h-l));
					break;
				}
			}
		}
	}
	cout << ans << endl;
	return 0;
}