#include <bits/stdc++.h>

#define rep(i, x, n) for (int i = x; i < n; i++)
typedef long long ll;

const int INF = 1e9 + 7;

using namespace std;

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

	int n;
	cin >> n;

	vector<pair<int, int>> a(n);
	vector<pair<int, int>> b(n);
	rep (i, 0, n)
		cin >> a[i].first >> a[i].second;
	rep (i, 0, n)
		cin >> b[i].first >> b[i].second;
	sort(a.begin(), a.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
			return a.second > b.second;
		});
	sort(b.begin(), b.end());

	int res = 0;
	vector<int> memo(n, 0);
	rep (i, 0, n) { // loop a
		rep (j, 0, n) {
			if (b[j].first > a[i].first && b[j].second > a[i].second && memo[j] == 0) {
				res++;
				memo[j] = 1;
				break;
			}
		}
	}

	cout << res << 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, x, n) for (int i = x; i < n; i++)
typedef long long ll;

const int INF = 1e9 + 7;

using namespace std;

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

	int n;
	cin >> n;

	vector<pair<int, int>> a(n);
	vector<pair<int, int>> b(n);
	rep (i, 0, n)
		cin >> a[i].first >> a[i].second;
	rep (i, 0, n)
		cin >> b[i].first >> b[i].second;
	sort(a.begin(), a.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
			return a.second > b.second;
		});
	sort(b.begin(), b.end());

	int res = 0;
	vector<int> memo(n, 0);
	rep (i, 0, n) { // loop a
		rep (j, 0, n) {
			if (b[j].first > a[i].first && b[j].second > a[i].second && memo[j] == 0) {
				res++;
				memo[j] = 1;
				break;
			}
		}
	}

	cout << res << endl;
	return 0;
}

```