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)

import java.util.*;
import java.io.*;

public class Main {
	static int a[], b[];
	static long cnt = 0;
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		a= new int[n];
		b= new int[n];
		
		for(int i=0;i<n;i++){
			a[i] = sc.nextInt();
		}
		for(int i=0;i<n;i++){
			b[i] = sc.nextInt();
		}
		
		int ans = 0;
		for(int i=1;i<=29;i++){
			int cntone = 0;
			int T = 1<<(i-1);
			int tmpb[] = new int[n];
			
			for(int j=0;j<n;j++){
				tmpb[j] = b[j]%(2*T);
			}
			
			Arrays.sort(tmpb);
			
			for(int j=0;j<n;j++){
				
				int l1 = binarySearch(tmpb, a[j]%(2*T), T);
				int r1 = binarySearch(tmpb, a[j]%(2*T), 2*T);
				
				int l2 = binarySearch(tmpb, a[j]%(2*T), 3*T);
				int r2 = binarySearch(tmpb, a[j]%(2*T), 4*T);
				
				cntone += (r1-l1)+(r2-l2);
			}
			
			if(cntone %2==1)
				ans |= 1<<(i-1);
		}
		System.out.println(ans);
	}
	static int binarySearch(int tmp[], int a, int T) {
		int l = 0, r = b.length;
		while(l<r){
			int mid=(l+r)/2;
			if(a+tmp[mid]>=T)	r = mid;
			else l = mid+1;
		}
		
		return l;
	}	
}