#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;

signed main(void)
{
	int n, cnt;
	cin >> n;
	vector<bool> red(n, true);
	vector<P> r(n), b(n);

	rep(i, n) cin >> r[i].first >> r[i].second;
	rep(i, n) cin >> b[i].first >> b[i].second;
	sort(r.begin(), r.end());
	sort(b.begin(), b.end());
	cnt = 0;
	rep(i, n)
	{
		int j = 0;
		vector<P> tmp;
		while(r[j].first < b[i].first && j < n)
		{
			if (red[j] && r[j].second < b[i].second)
				tmp.push_back(P(r[j].second, j));
			j++;
		}
		if (tmp.size() != 0)
		{
			sort(tmp.rbegin(), tmp.rend());
			red[tmp[0].second] = false;
			cnt++;
		}
	}
	cout << cnt << endl;
	return 0;
}

### Prompt

Create a solution in cpp for 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>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;

signed main(void)
{
	int n, cnt;
	cin >> n;
	vector<bool> red(n, true);
	vector<P> r(n), b(n);

	rep(i, n) cin >> r[i].first >> r[i].second;
	rep(i, n) cin >> b[i].first >> b[i].second;
	sort(r.begin(), r.end());
	sort(b.begin(), b.end());
	cnt = 0;
	rep(i, n)
	{
		int j = 0;
		vector<P> tmp;
		while(r[j].first < b[i].first && j < n)
		{
			if (red[j] && r[j].second < b[i].second)
				tmp.push_back(P(r[j].second, j));
			j++;
		}
		if (tmp.size() != 0)
		{
			sort(tmp.rbegin(), tmp.rend());
			red[tmp[0].second] = false;
			cnt++;
		}
	}
	cout << cnt << endl;
	return 0;
}

```