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 Python (3.4.3) (judge verdict: Accepted)

N = int(input())

"""考え方
1.青色を第一要素（x）が小さい順にソートする
2.青色の点より左にある赤い点のうち、yが小さく、かつ一番大きい赤い点を選択する
"""

## 1.必要な配列を取得し、x軸でソートしておく
## 1-1.赤色
AB = []
for n in range(N):
  AB.append(list(map(int, input().split())))
  
AB = sorted(AB, key=lambda x: x[1], reverse=True)

## 1-2.青色
CD = []
for n in range(N):
  CD.append(list(map(int, input().split())))
CD = sorted(CD, key=lambda x: x[0], reverse=False)


## 2.青色の点について、左から順に取得していく
used_idx = []
ans = 0
for _ in CD:
  c = _[0]
  d = _[1]
  red_max_y = -1
  ## 左側にある赤色の要素を検索
  for idx, __ in enumerate(AB):
    a = __[0]
    b = __[1]
    if a < c:
      if b < d:
        if idx not in used_idx:
          used_idx.append(idx)
          ans += 1
          break
  
    
print(ans)