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

int n, x, y, ans, k = -1;
bool f[101];
pair<int, int> R[101], B[101];
int main() {
	cin >> n;
	rep(i, n) {
		cin >> x >> y;
		R[i] = make_pair(x, y);
	}
	rep(i, n) {
		cin >> x >> y;
		B[i] = make_pair(x, y);
	}
	sort(R, R + n);
	sort(B, B + n);
	rep(i, n) {
		rep(j, n) {
			if (B[i].first > R[j].first && B[i].second > R[j].second) {
				if (f[j] == 0) {
					if (k == -1) k = j;
					else if (R[k].second < R[j].second) k = j;
				}
			}
		}
		if (k != -1) {
			f[k] = 1;
			ans++;
		}
		k = -1;
	}
	cout << ans << endl;
	return 0;
}
### 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 <bits/stdc++.h>
#define rep(i, n) for(int i = 0;i < n;i++)
using namespace std;

int n, x, y, ans, k = -1;
bool f[101];
pair<int, int> R[101], B[101];
int main() {
	cin >> n;
	rep(i, n) {
		cin >> x >> y;
		R[i] = make_pair(x, y);
	}
	rep(i, n) {
		cin >> x >> y;
		B[i] = make_pair(x, y);
	}
	sort(R, R + n);
	sort(B, B + n);
	rep(i, n) {
		rep(j, n) {
			if (B[i].first > R[j].first && B[i].second > R[j].second) {
				if (f[j] == 0) {
					if (k == -1) k = j;
					else if (R[k].second < R[j].second) k = j;
				}
			}
		}
		if (k != -1) {
			f[k] = 1;
			ans++;
		}
		k = -1;
	}
	cout << ans << endl;
	return 0;
}
```