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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N = $stdin.gets.chomp.to_i
ab = []
cd = []
N.times do |i|
    ab[i] = $stdin.gets.chomp.split(' ').map(&:to_i)
end
N.times do |i|
    cd[i] = $stdin.gets.chomp.split(' ').map(&:to_i)
end

ab = ab.sort_by{|p| -(p[0] + p[1])}

def count(ab_arg, cd_arg)
    return 0 if ab_arg.length == 0
    ab = ab_arg.dup
    cd = cd_arg.dup
    ab_i = 0
    candidates = []
    ab.length.times do |i|
        candidates = cd.select{|p| p[0] > ab[i][0] && p[1] > ab[i][1]}
        if candidates.length > 0
            ab_i = i
            break
        end
    end

    max = 0
    ab.delete_at(ab_i)
    candidates.each do |c|
        cd2 = cd.dup
        cd2.delete(c)
        max = [max, count(ab, cd2) + 1].max
    end
    return max
end

print count(ab, cd)