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>
#define int llint
#define rep(i, a, n) for(int (i) = (int)(a);(i) < (int)(n);++(i))
#define repe(i, a, n) for(int (i) = (int)(a);(i) <= (int)(n);++(i))
#define repif(i, a, x) for(int (i) = (int)(a);(x);++(i))
#define repr(i, a, n) for(int (i) = ((int)(a)-1);(i) >= (int)(n);--(i))
#define reper(i, a, n) for(int (i) = (int)(a);(i) >= (int)(n);--(i))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define SIZE(x) ((int)(x).size())
#define ALL(x) (x).begin(), (x).end()
#define RNG(x, a, n) &((x)[a]), &((x)[n])
#define ADD(a, b) ((((a)%MOD)+((b)%MOD))%MOD)
#define MUL(a, b) ((((a)%MOD)*((b)%MOD))%MOD)
#define CEIL(x) ((int)ceil(((double)x)))
#define POW(x, y) ((int)pow(x, y))
#define UNIQUE(x) (x).erase(unique(ALL((x))), (x).end())
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define fcout cout << fixed << setprecision(10)
#define pb push_back
#define pf push_front
#define mp make_pair
#define mt make_tuple
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define EPS (1e-20)
#define INF 2147483647
#define MOD 1000000007
using llint = long long;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using SI = std::set<int>;
using MII = std::map<int, int>;
using TIII = std::tuple<int, int, int>;
using VLL = std::vector<llint>;
using MCI = std::map<char, int>;
using namespace std;


int N, K;
int x[60], y[60];


int cnt(int x1, int x2, int y1, int y2){
	int res = 0;
	rep(i, 0, N)
		if(x1 <= x[i] && x[i] <= x2 && y1 <= y[i] && y[i] <= y2) res++;
	return res;
}


int solve(){
	int ans = LLONG_MAX;
	rep(i, 0, N){
		rep(j, i+1, N){
			rep(k, 0, N){
				rep(l, k+1, N){
					int x1 = min(x[i], x[j]);
					int x2 = max(x[i], x[j]);
					int y1 = min(y[k], y[l]);
					int y2 = max(y[k], y[l]);
					if(cnt(x1, x2, y1, y2) >= K) chmin(ans, (x2-x1)*(y2-y1));
				}
			}
		}
	}
	return ans;
}



signed main(){
	cin >> N >> K;
	rep(i, 0, N) cin >> x[i] >> y[i];

	cout << solve() << endl;

	return 0;
}