Problem:
Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points.
The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

All input values are integers.

1 \leq N \leq 100

0 \leq a_i, b_i, c_i, d_i < 2N

a_1, a_2, ..., a_N, c_1, c_2, ..., c_N are all different.

b_1, b_2, ..., b_N, d_1, d_2, ..., d_N are all different.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_N b_N
c_1 d_1
c_2 d_2
:
c_N d_N

Output

Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

/*
String[]の初期化、整数の型がlongになっているか
関数（splitとかcontainsとか）内では"+"とかは正規表現となされるので"\\+"とする
配列のコピーはarr1=arr.clone()
HashSet<>[]はおのおの初期化した？？？？？
 */
import java.applet.Applet;
import java.util.*;
import java.awt.*;
import java.awt.Graphics.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        int n = sc.nextInt();
        int inf=Integer.MAX_VALUE;
        Point[] red=new Point[n];
        for (int i=0;i<n;i++)red[i]=new Point(sc.nextInt(),sc.nextInt());
        Point[] blue=new Point[n];
        for (int i=0;i<n;i++)blue[i]=new Point(sc.nextInt(),sc.nextInt());
        Arrays.sort(blue,(p1,p2)->p1.x-p2.x);
        int ans=0;
        for (int i = 0; i < n; i++) {
            Point b=blue[i];
            int red_y=-1;
            int red_index=-1;
            for (int j = 0; j < n; j++) {
                Point r=red[j];
                if(d(b,r)){
                    if(red_y<r.y){
                        red_index=j;
                    }
                }
            }
            if(red_index>-1){
                if(d(b,red[red_index]))ans++;
                red[red_index]=new Point(inf,inf);
            }
        }
        out.println(ans);
    }
    static boolean d(Point blue,Point red){
        return red.x<blue.x&&red.y<blue.y;
    }
}