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: Wrong Answer)

import sys
input = sys.stdin.readline
iN = int(input())
aR = [(i,[int(_) for _ in input().split()]) for i in range(iN)]
aB = [(i,[int(_) for _ in input().split()]) for i in range(iN)]

dBR={} #青→赤 青からみた組の可能性
dRB={} #赤←青 赤からみた組の可能性
dBRc={}
for i,eB in aB:
    for j,eR in aR:
        if eB[0] > eR[0] and eB[1] > eR[1]:
            if j in dRB:
                dRB[j].add(i)
            else:
                dRB[j]={i}
            if i in dBR:
                dBR[i].add(j)
                dBRc[i] += 1
            else:
                dBR[i] = {j}
                dBRc[i] = 1

#組み合わせの数が両方少ない方から決めていく
dDicided = set()
for k,v in sorted(dBRc.items(),key=lambda x:x[1]):
    for r, lr in  sorted(map(lambda r :[r,len(dRB[r])], dBR[k]),key=lambda x:x[1]) :
        if r not in dDicided :
            dDicided.add(r)
            break
print(len(dDicided))