#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
#define int long long
#define F first
#define S second
signed main() {
	int n, ans = 0; cin >> n;
	pair<int, int>r[110], b[110];
	bool r_[110];

	for (int i = 0; i < n; i++)cin >> r[i].F >> r[i].S; sort(r, r + n);
	for (int i = 0; i < n; i++)cin >> b[i].F >> b[i].S; sort(b, b + n);
	for (int i = 0; i < n; i++)r_[i] = 0;

	for (int i = 0; i < n; i++) {
		int k = -1, j = 0;
		for (int j = 0; j < n&&r[j].F < b[i].F; j++) {
			if (!r_[j] && r[j].S < b[i].S) {
				if (k < 0)k = j;
				else {
					if (r[k].S < r[j].S)k = j;
				}
			}
		}
		if (k >= 0) {
			r_[k] = 1; ans++;
		}
	}
	cout << ans << endl;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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.

Examples

Input

3
2 0
3 1
1 3
4 2
0 4
5 5


Output

2


Input

3
0 0
1 1
5 2
2 3
3 4
4 5


Output

2


Input

2
2 2
3 3
0 0
1 1


Output

0


Input

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7


Output

5


Input

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9


Output

4

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
#define int long long
#define F first
#define S second
signed main() {
	int n, ans = 0; cin >> n;
	pair<int, int>r[110], b[110];
	bool r_[110];

	for (int i = 0; i < n; i++)cin >> r[i].F >> r[i].S; sort(r, r + n);
	for (int i = 0; i < n; i++)cin >> b[i].F >> b[i].S; sort(b, b + n);
	for (int i = 0; i < n; i++)r_[i] = 0;

	for (int i = 0; i < n; i++) {
		int k = -1, j = 0;
		for (int j = 0; j < n&&r[j].F < b[i].F; j++) {
			if (!r_[j] && r[j].S < b[i].S) {
				if (k < 0)k = j;
				else {
					if (r[k].S < r[j].S)k = j;
				}
			}
		}
		if (k >= 0) {
			r_[k] = 1; ans++;
		}
	}
	cout << ans << endl;
}
```