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 C# (Mono-mcs 6.8.0.105) (judge verdict: Accepted)

using System;
using System.Collections;
using System.IO;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Collections.Generic;
using System.Threading.Tasks;

class Program
{
	//Main
	static void Main(string[] args)
	{
		//読み込む
		int N = int.Parse(Console.ReadLine());

		//x座標とy座標を合わせたジャグ配列
		//ソートの都合上、{y,x}とする
		int[][] redCoord = new int[N][];
		//ソート の都合上、{x,y}とする
		int[][] blueCoord = new int[N][];

		//読み込む
		for(int i=0; i<N; i++){
			string[] S = Console.ReadLine().Split(' ');
			int a = int.Parse(S[0]);
			int b = int.Parse(S[1]);

			redCoord[i] = new int[2]{b,a};
		}
		for(int i=0; i<N; i++){
			string[] S = Console.ReadLine().Split(' ');
			int c = int.Parse(S[0]);
			int d = int.Parse(S[1]);

			blueCoord[i] = new int[2]{c,d};
		}

		//座標の順に並び替え
		Array.Sort(redCoord,StructuralComparisons.StructuralComparer);
		Array.Sort(blueCoord,StructuralComparisons.StructuralComparer);

		/*debug
		for(int i=0; i<N; i++){
			Console.WriteLine("red:y={0},x={1} | blue:x={2},y={3}",redCoord[i][0],redCoord[i][1],
			blueCoord[i][0],blueCoord[i][1]);
		}
		*/

		//既に選ばれたペアの数
		int ans = 0;

		//各赤について、既にペアがいるかを示す
		bool[] selectedRed = new bool[N];

		//貪欲に、各青が一番ギリギリの赤を取っていく
		for(int i=0; i<N; i++){
			for(int j=N-1; j>=0; j--){
				//既に選ばれている場合、スキップ
				if(selectedRed[j]){
					continue;
				}
				//ペアになれる場合ペアにして、次の青へ
				/*
				redとblueでxyが逆なので注意
				*/
				else if(redCoord[j][1] < blueCoord[i][0] && redCoord[j][0] < blueCoord[i][1]){
						ans ++;
						selectedRed[j] = true;
						//Console.WriteLine("red:{0},blue{1}",j,i);
						break;
				}
			}
		}

		Console.WriteLine(ans);
	}
}