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 Rust (1.42.0) (judge verdict: Accepted)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num::clamp;
use proconio::{input, marker::*, fastout};
use superslice::*;

#[fastout()]
fn main() {
    input! {
        n: usize,
        red: [(usize, usize); n],
        mut blue: [(usize, usize); n]
    }
    blue.sort();

    let mut b2r = vec![Vec::new(); n];
    for i in 0..n {
        let (bx, by) = blue[i];
        for j in 0..n {
            let (rx, ry) = red[j];
            if rx < bx && ry < by {
                b2r[i].push(red[j]);
            }
        }
        b2r[i].sort_by_key(|&(_, ry)| Reverse(ry));
    }

    let mut red_set = red.into_iter().collect::<HashSet<_>>();
    let mut ans = 0;
    for red_vec in b2r {
        for red in &red_vec {
            if red_set.contains(red) {
                ans += 1;
                red_set.remove(red);
                break;
            }
        }
    }
    println!("{}", ans);
}