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)

import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)


class BipartiteMatching:
    def __init__(self, N, M):
        self.N = N
        self.M = M
        self.pairA = [-1] * N
        self.pairB = [-1] * M
        self.edge = [[] for _ in range(N)]
        self.was = [0] * N
        self.iter = 0

    def add(self, s, t):
        self.edge[s].append(t)

    def _DFS(self, s):
        self.was[s] = self.iter
        for t in self.edge[s]:
            if self.pairB[t] == -1:
                self.pairA[s] = t
                self.pairB[t] = s
                return True
        for t in self.edge[s]:
            if self.was[self.pairB[t]] != self.iter and self._DFS(self.pairB[t]):
                self.pairA[s] = t
                self.pairB[t] = s
                return True
        return False

    def solve(self):
        res = 0
        while True:
            self.iter += 1
            found = 0
            for i in range(self.N):
                if self.pairA[i] == -1 and self._DFS(i):
                    found += 1
            if not found:
                break
            res += found
        return res


if __name__ == "__main__":
    N = int(input())
    AB = tuple(tuple(map(int, input().split())) for _ in range(N))
    CD = tuple(tuple(map(int, input().split())) for _ in range(N))
    match = BipartiteMatching(N + 5, N + 5)

    for i, (a, b) in enumerate(AB):
        for j, (c, d) in enumerate(CD):
            if a < c and b < d:
                match.add(i, j)
    print(match.solve())