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;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = (A); I < (B); ++I)
#define FORR(I,A,B) for(ll I = ((B)-1); I >= (A); --I)
#define TO(x,t,f) ((x)?(t):(f))
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd(ll a,ll b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);}
ll lcm(ll a,ll b){ll c=gcd(a,b);return ((a/c)*(b/c)*c);}//saisyo kobaisu
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=LLONG_MAX;
const ll MOD=1e9+7;
#define pri(a) cout << (a) << endl



int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll N,K;
	cin >> N >> K;
	vector<ll> x(N),y(N);
	FOR(i,0,N){
		cin >> x[i] >> y[i];
	}
	ll ans = INF;
	FOR(i,0,N){
		FOR(j,0,N){
			FOR(k,0,N){
				FOR(l,0,N){
					ll yu,yd,xr,xl;
					yu = max( {y[i],y[j],y[k],y[l]} );
					yd = min( {y[i],y[j],y[k],y[l]} );
					xr = max( {x[i],x[j],x[k],x[l]} );
					xl = min( {x[i],x[j],x[k],x[l]} );
					if((xr-xl)*(yu-yd)==0)continue;
					ll n = 0;
					FOR(m,0,N){
						if(xl<=x[m] && x[m]<=xr)
							if(yd<=y[m] && y[m]<=yu)
								n++;
						if(n>=K)
							ans = min(ans,(xr-xl)*(yu-yd));
					}
				}
			}
		}
	}
	cout << ans << endl;
}