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 <string>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <numeric>
#include <regex>
#include <tuple>
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
#define modulo 1000000007 // 10^9 + 7
#define INF 1000000000 // 10^9

int x[50];
int y[50];
int N, K;

void input() {
	cin >> N >> K;
	for (int i = 0; i < N; i++) cin >> x[i] >> y[i];
}

void solve() {
	ll ans = INF;
	ll x1, x2, y1, y2;

	for (int i = 0; i < N; i++) { // x
		for (int j = 0; j < N; j++) { // x
			for (int k = 0; k < N; k++) { // y
				for (int l = 0; l < N; l++) { // y
					x1 = min(x[i], x[j]);
					x2 = max(x[i], x[j]);
					y1 = min(y[k], y[l]);
					y2 = max(y[k], y[l]);
					int cnt = 0;

					for (int m = 0; m < N; m++) if (x1 <= x[m] && x[m] <= x2 && y1 <= y[m] && y[m] <= y2) cnt++;
					if (cnt >= K) ans = min(ans, (x2 - x1)*(y2 - y1));
				}
			}
		}
	}
	
	cout << ans << endl;
}

int main() {
	input();
	solve();

	return 0;
}