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 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections;
using System.Collections.Generic;
class Program
{
  public static int N;
  public static List<int>[] list = new List<int>[110];

  static void Main(string[] args)
  {
    int[] rx = new int[110];
    int[] ry = new int[110];
    int[] bx = new int[110];
    int[] by = new int[110];
    // スペース区切りの整数の入力
    N = int.Parse(Console.ReadLine());
    for (int i = 0; i < N; i++) {
      string[] input = Console.ReadLine().Split(' ');
      rx[i] = int.Parse(input[0]);
      ry[i] = int.Parse(input[0]);
    }
    for (int i = 0; i < N; i++) {
      string[] input = Console.ReadLine().Split(' ');
      bx[i] = int.Parse(input[0]);
      by[i] = int.Parse(input[0]);
    }

    int lastvalue = 0;

    for (int j = 0; j < N; j++) { //ソート
      for (int i = 1; i < N; i++) {
        if (bx[i] + by[i] < bx[i - 1] + by[i - 1]) {
          int tmp = bx[i - 1];
          bx[i - 1] = bx[i];
          bx[i] = tmp;
          tmp = by[i - 1];
          by[i - 1] = by[i];
          by[i] = tmp;
        }
      }
    }

    for (int i = 0; i < N; i++) { //リスト追加
      list[i] = new List<int>();
      for (int j = 0; j < N; j++) {
        if (rx[j] < bx[i] && ry[j] < by[i] ) {
          list[i].Add(j);
        }
      }
      list[i].Add(-1);
    }

    for(int i=N-1;i>0;i--){
      if (list[i].Count == 0)N--;
      if (list[i].Count != 0)break;;
    }


    search( 0);
    Console.WriteLine(maxv);
  }

  public static int ctr=0;
  public static bool[] chk = new bool[110];
  public static int maxv=0;
  public static int[] lastAdd = new int[110];

  public static void search( int b ) {
    if (ctr == N) {
      if (lastAdd[ctr-1]!=-1)chk[lastAdd[ctr-1]] = false;
      if (maxv<b)maxv=b;
      ctr--;
      return;
    }

    for (int i = 0; i < list[ctr].Count; i++) {
      if (list[ctr][i] != -1 && !chk[list[ctr][i]] ) {
        lastAdd[ctr] = list[ctr][i];
        chk[list[ctr][i]] = true;
        ctr++;
        search( b + 1);
      }else{
        lastAdd[ctr] = -1;
        ctr++;
        search( b );
      }
    }
    if (ctr>0)if (lastAdd[ctr-1]!=-1)chk[lastAdd[ctr-1]] = false;
    ctr--;
    return;
  }
  

}