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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vb = vector<bool>;
using vs = vector<string>;
using msi = map<string, int>;
using mii = map<int, int>;
using psi = pair<string, int>;
using pii = pair<int, int>;
using vlai = valarray<int>;
#define rep(i,n) for(int i=0;i<n;i++)
#define range(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define INF 1E9
#define LINF 1E17
#define EPS 1E-9
#define MOD 1000000007
#define PI 3.1415926535897932384

template<class S, class T>ostream& operator<<(ostream&os, pair<S, T>p) { os << "[" << p.first << ", " << p.second << "]"; return os; };
template<class S>auto&operator<<(ostream&os, vector<S>t) { bool a = 1; for (auto s : t) { os << (a ? "" : " ") << s; a = 0; } return os; }

//int dx[]={1,1,1,0,-1,-1,-1,0},dy[8]={1,0,-1,-1,-1,0,1,1,1};
//
constexpr ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }

int main() {
	int n, K, in1, in2;
	vi x, y;
	cin >> n >> K;
	rep(i, n) {
		cin >> in1 >> in2;
		x.push_back(in1);
		y.push_back(in2);
	}
	sort(all(x)); sort(all(y));
	ll ans = LINF;
	rep(i, n) {
		range(j, i + 1, n) {
			rep(k, n) {
				range(l, k + 1, n) {
					int count = 0;
					rep(m, n) {
						if (x[i] <= x[m] && x[m] <= x[j] && y[k] <= y[m] && y[m] <= y[l]) {
							count++;
						}
					}
					if (count >= K) {
						ll area = (ll)(x[j] - x[i])*(y[l] - y[k]);
						ans = min(ans, area);
					}
				}
			}
		}
	}
	cout << ans << endl;
}