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.*;
import java.io.*;
 
public class Main {
    public static void main(String[] args) throws Exception {
        FastScanner sc = new FastScanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        long[] xl = new long[n];
        long[] yl = new long[n];
        ArrayList<Point> pl = new ArrayList<>();
        for(int i = 0; i < n; i++){
            long x = sc.nextLong();
            long y = sc.nextLong();
            xl[i] = x;
            yl[i] = y;
            pl.add(new Point(x,y));
        }
        Arrays.sort(xl);
        Arrays.sort(yl);
        long ans = Long.MAX_VALUE;
        for(int i = 0; i < n; i++){
            for(int i2 = i+1; i2 < n; i2++){
                for(int j = 0; j < n; j++){
                    for(int j2 = j+1; j2 < n; j2++){
                        long lx = xl[i];
                        long rx = xl[i2];
                        long ly = yl[j];
                        long ry = yl[j2];
                        long S = Math.abs(rx-lx)*Math.abs(ry-ly);
                        if(ans <= S){
                            continue;
                        }
                        int cnt = 0;
                        for(int m = 0; m < n; m++){
                            Point p = pl.get(m);
                            if(lx <= p.x && p.x <= rx && ly <= p.y && p.y <= ry) cnt++;   
                        }
                        if(cnt >= k){
                            ans = S;
                        }
                    }
                }
            }
        }
        System.out.println(ans);
    }
    
}

class Point{
    long x,y;
    public Point(long x, long y){
        this.x = x;
        this.y = y;
    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;
    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public double nextDouble() {
         return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    } 
}