#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N; cin >> N;

	P r[200], b[200];

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

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

	sort(r, r+N); sort(b, b+N);

	int ans = 0;

	for (int i = 0;i < N;i++){
		int highest = -1;
		int idx = -1;
		for (int j = 0;j < N;j++){
			if(r[j].first <= b[i].first && r[j].second <= b[i].second){
				if(highest < r[j].second){
					highest = r[j].second;
					idx = j;
				}
			}
		}
		if(highest != -1){
			ans++;
			r[idx] = P(INT_MAX, INT_MAX);
		}
	}	

	cout << ans << endl;

	return 0;

}

### Prompt

Please create a solution in CPP 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 long long ll;
typedef pair<int,int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N; cin >> N;

	P r[200], b[200];

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

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

	sort(r, r+N); sort(b, b+N);

	int ans = 0;

	for (int i = 0;i < N;i++){
		int highest = -1;
		int idx = -1;
		for (int j = 0;j < N;j++){
			if(r[j].first <= b[i].first && r[j].second <= b[i].second){
				if(highest < r[j].second){
					highest = r[j].second;
					idx = j;
				}
			}
		}
		if(highest != -1){
			ans++;
			r[idx] = P(INT_MAX, INT_MAX);
		}
	}	

	cout << ans << endl;

	return 0;

}

```