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 OCaml (4.10.0) (judge verdict: Accepted)

let n = read_int()

let ab = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun a b -> a, b))

let cd = Array.init n (fun _ -> Scanf.scanf " %d %d" (fun c d -> c, d))

let list = Array.make n 0
let li = ref 0

let rec f i j p =
  if i = n then p
  else
    let c, d = cd.(i) in
    let rec g j =
      if j < n then (
        let a, b = ab.(j) in
        if a < c then (
          list.(!li) <- b;
          li := !li + 1;
          g (j + 1)
        ) else j
      ) else j in
    let j = g j in
    let rec g k l m =
      if k = !li then l
      else if list.(k) < d && m < list.(k) then g (k + 1) k list.(k)
      else g (k + 1) l m in
    let l = g 0 ~-1 ~-1 in
    if l > -1 then (
      list.(l) <- list.(!li - 1);
      li := !li - 1;
      f (i + 1) j (p + 1)
    ) else f (i + 1) j p

let () =
  Array.sort (fun (x1, _) (x2, _) -> x1 - x2) ab;
  Array.sort (fun (x1, _) (x2, _) -> x1 - x2) cd;
  let ans = f 0 0 0 in
  Printf.printf "%d\n" ans