#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;
int N;
vector<P> red, blue;
int ans = 0;

int main() {
	cin >> N;
	for (int i = 0; i < 2*N; i++) {
		int x, y;
		cin >> x >> y;
		if (i < N) {
			red.push_back(P(x, y));
		}
		else {
			blue.push_back(P(x, y));
		}
	}

	sort(red.begin(), red.end());
	for (int i = N - 1; i >= 0; i--) {
		int rx = red[i].first, ry = red[i].second;
		int min_by = 200, buf = -1;
		for (int j = 0; j < (int) blue.size(); j++) {
			int bx = blue[j].first, by = blue[j].second;
			if (bx > rx && by > ry && min_by > by) {
				min_by = by;
				buf = j;
			}
		}

		if (buf != -1) {
			blue.erase(blue.begin() + buf);
			ans++;
		}
	}

	cout << ans << endl;
}
### Prompt

Your task is to create a Cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;
int N;
vector<P> red, blue;
int ans = 0;

int main() {
	cin >> N;
	for (int i = 0; i < 2*N; i++) {
		int x, y;
		cin >> x >> y;
		if (i < N) {
			red.push_back(P(x, y));
		}
		else {
			blue.push_back(P(x, y));
		}
	}

	sort(red.begin(), red.end());
	for (int i = N - 1; i >= 0; i--) {
		int rx = red[i].first, ry = red[i].second;
		int min_by = 200, buf = -1;
		for (int j = 0; j < (int) blue.size(); j++) {
			int bx = blue[j].first, by = blue[j].second;
			if (bx > rx && by > ry && min_by > by) {
				min_by = by;
				buf = j;
			}
		}

		if (buf != -1) {
			blue.erase(blue.begin() + buf);
			ans++;
		}
	}

	cout << ans << endl;
}
```