#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

typedef pair<int, int> P;

int main() {
	int N;
	cin >> N;

	vector<P> f(N);
	vector<P> s(N);

	for (int i = 0; i < N; i++) {
		int a, b;
		cin >> a >> b;
		f[i].first = b;
		f[i].second = a;
	}

	for (int i = 0; i < N; i++) {
		int c, d;
		cin >> c >> d;
		s[i].first = c;
		s[i].second = d;
	}
	
    sort(f.begin(),f.end());
    sort(s.begin(),s.end());

	int count = 0;
	for (int i = 0; i < N; i++) {
		for (int j = f.size()-1; j >=0; j--) {
			if (s[i].first >= f[j].second && s[i].second >= f[j].first) {
				count++;
				f.erase(f.begin() + j);
				break;
			}
		}
	}

	cout << count << endl;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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<vector>
#include<algorithm>
using namespace std;

typedef pair<int, int> P;

int main() {
	int N;
	cin >> N;

	vector<P> f(N);
	vector<P> s(N);

	for (int i = 0; i < N; i++) {
		int a, b;
		cin >> a >> b;
		f[i].first = b;
		f[i].second = a;
	}

	for (int i = 0; i < N; i++) {
		int c, d;
		cin >> c >> d;
		s[i].first = c;
		s[i].second = d;
	}
	
    sort(f.begin(),f.end());
    sort(s.begin(),s.end());

	int count = 0;
	for (int i = 0; i < N; i++) {
		for (int j = f.size()-1; j >=0; j--) {
			if (s[i].first >= f[j].second && s[i].second >= f[j].first) {
				count++;
				f.erase(f.begin() + j);
				break;
			}
		}
	}

	cout << count << endl;
}
```