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

const int N = 1100;

struct Point {
	int x, y;
	Point(int x = 0, int y = 0) : x(x), y(y) {}
} P[N], Q[N];

int n;
int b[N], vis[N];

bool Find(int u) {
	for (int i = 1; i <= n; i++) if (P[u].x < Q[i].x && P[u].y < Q[i].y && !vis[i]) {
		vis[i] = 1;
		if (b[i] == -1 || Find(b[i])) {
			b[i] = u;
			return true;
		}
	}
	return false;
}

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> P[i].x >> P[i].y;
	for (int i = 1; i <= n; i++) cin >> Q[i].x >> Q[i].y;
	
	memset(b, 0xff, sizeof b);
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) vis[j] = 0;
		if (Find(i)) ans++;
	}
	cout << ans << "\n";
	
	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>
using namespace std;

const int N = 1100;

struct Point {
	int x, y;
	Point(int x = 0, int y = 0) : x(x), y(y) {}
} P[N], Q[N];

int n;
int b[N], vis[N];

bool Find(int u) {
	for (int i = 1; i <= n; i++) if (P[u].x < Q[i].x && P[u].y < Q[i].y && !vis[i]) {
		vis[i] = 1;
		if (b[i] == -1 || Find(b[i])) {
			b[i] = u;
			return true;
		}
	}
	return false;
}

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	cin >> n;
	for (int i = 1; i <= n; i++) cin >> P[i].x >> P[i].y;
	for (int i = 1; i <= n; i++) cin >> Q[i].x >> Q[i].y;
	
	memset(b, 0xff, sizeof b);
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) vis[j] = 0;
		if (Find(i)) ans++;
	}
	cout << ans << "\n";
	
	return 0;
}

```