#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int n,k,ans,di[101010];
struct d
{
	int x,y;
}a[10101],b[10101];
bool cmp(d u,d v){ return u.x<v.x;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&b[i].x,&b[i].y);
	sort(a+1,a+n+1,cmp);
	sort(b+1,b+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		while(a[k].x<=b[i].x&&k<=n) k++;
		int ma=0;
		for(int j=1;j<k;j++)
		if(!di[j]&&a[j].y<=b[i].y)
			ma=a[ma].y>a[j].y?ma:j;
	//	cout<<k<<" "<<ma<<endl;
		if(ma) di[ma]=1,ans++;
	}
	printf("%d",ans);
	return 0;
}
		
		
### Prompt

Please provide a Cpp coded solution to the problem described below:

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<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int n,k,ans,di[101010];
struct d
{
	int x,y;
}a[10101],b[10101];
bool cmp(d u,d v){ return u.x<v.x;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&a[i].x,&a[i].y);
	for(int i=1;i<=n;i++)
		scanf("%d%d",&b[i].x,&b[i].y);
	sort(a+1,a+n+1,cmp);
	sort(b+1,b+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		while(a[k].x<=b[i].x&&k<=n) k++;
		int ma=0;
		for(int j=1;j<k;j++)
		if(!di[j]&&a[j].y<=b[i].y)
			ma=a[ma].y>a[j].y?ma:j;
	//	cout<<k<<" "<<ma<<endl;
		if(ma) di[ma]=1,ans++;
	}
	printf("%d",ans);
	return 0;
}
		
		
```