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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>

using namespace std;

int main() {
    const int red = 0, blue = 1;

    int n;
    cin >> n;

    vector<int> y(2 * n, 2 * n), clr(2 * n);
    for (int i = 0; i < n; i++) {
        int a, b;
        cin >> a >> b;
        y[a] = b;
        clr[a] = red;
    }

    for (int i = 0; i < n; i++) {
        int c, d;
        cin >> c >> d;
        y[c] = d;
        clr[c] = blue;
    }

    int res = 0;
    for (int i = 0; i < 2 * n; i++) {
        if (clr[i] == blue) {
            int x = -1, h = -1;
            for (int j = 0; j < i; j++) {
                if (clr[j] == red && h < y[j] && y[j] < y[i]) {
                    x = j;
                    h = y[j];
                }
            }

            if (x != -1) {
                y[x] = 2 * n;
                res++;
            }
        }
    }

    cout << res << endl;
}