Problem: 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.

Examples

Input

4 4
1 4
3 3
6 2
8 1


Output

21


Input

4 2
0 0
1 1
2 2
3 3


Output

1


Input

4 3
-1000000000 -1000000000
1000000000 1000000000
-999999999 999999999
999999999 -999999999


Output

3999999996000000001

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		long[] x = new long[N];
		long[] y = new long[N];
		for(int i=0; i<N; i++) {
			x[i] = sc.nextLong();
			y[i] = sc.nextLong();
		}
		long ans = Long.MAX_VALUE;
		for(int i=0; i<N; i++) {
			for(int j=i+1; j<N; j++) {
				for(int k=0; k<N; k++) {
					for(int l=k+1; l<N; l++) {
						long p1 = Math.min(x[i], x[j]);
						long p2 = Math.max(x[i], x[j]);
						long p3 = Math.min(y[k], y[l]);
						long p4 = Math.max(y[k], y[l]);
						int count = 0;
						for(int h=0; h<N; h++) {
							if(p1<=x[h] && x[h]<=p2 && p3<=y[h] && y[h] <= p4) count++;
						}
						if(count>=K) {
							ans = Math.min(ans, (p2-p1)*(p4-p3));
						}
					}
				}
			}
		}
		System.out.println(ans);
	}
	
	// print
	static void print(String s) {
		System.out.println(s);
	}
	
	// union find lib
	// usage:
	// 最初にinitを呼ぶ
	// root: 直接は呼ばないで
	// unite: まとめる
	// same: グループ判定
	static void init(int par[], int N) {
		for(int i=0; i<N; i++) {
			par[i] = i;
		}
	}
	
	static int root(int x, int [] par) {
		if(par[x] == x) {
			return x;
		} else {
			return (par[x] = root(par[x], par));
		}
	}
	
	static boolean same(int x, int y, int[] par) {
		return root(x, par) == root(y, par);
	}
	
	static void unite(int x, int y, int[] par) {
		x = root(x, par);
		y = root(y, par);
		if(x == y) return;
		par[x] = y;
	}
	
	// end union find lib
	
	// number lib
	
	static int max(int...ls) {
		int ans = Integer.MIN_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.max(ans, ls[i]);
		}
		return ans;
	}
	
	static int min(int...ls) {
		int ans = Integer.MAX_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.min(ans, ls[i]);
		}
		return ans;
	}
	
	static long lcm(long a, long b) {
		return a*(b/gcd(a, b));
	}
	
	static long gcd(long a, long b) {
		long ta = Math.max(a, b);
		long tb = Math.min(a, b);
		long tmp;
		while((tmp = ta%tb) != 0) {
			ta = tb;
			tb = tmp;
		}
		return tb;
	}
	
	static long modcomb(long n, long k, long mod) {
		if(k==1) {
			return n;
		}
		
		long ans = 1;
		for(long i=n; i>=n-k+1; i--) {
			ans = (ans * i)%mod;
		}
		for(long i=k; 0<i; i--) {
			ans = (ans * modpow(i, mod-2, mod)) % mod;
		}
		return ans;
	}
	
	static long modpow(long a, long b, long mod) {
		if(b==0) return 1;
		if(b%2==0) {
			long d = modpow(a, b/2, mod);
			return (d*d)%mod;
		} else {
			return (a*modpow(a, b-1, mod))%mod;
		}
	}
	
	static int disit(long a, long d) {
		int count = 0;
		while(a!=0) {
			a = a/d;
			count++;
		}
		return count;
	}
	
	// end number lib
}