#include<bits/stdc++.h>
using namespace std;
struct point{
	int x;
	int y;
	int used;
}R[105],B[105];
bool cmp(point a,point b)
{
    return a.x<b.x;
}
int main()
{
	int n;
	int a,b;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d %d",&R[i].x,&R[i].y);
	for(int i=0;i<n;i++)
		scanf("%d %d",&B[i].x,&B[i].y); 
	sort(B,B+n,cmp);
	int cnt=0;
	for(int i=0;i<n;i++)
	{
		int miny=200;
		int y=-1;
		for(int j=0;j<n;j++)
		{
			if(B[i].x>R[j].x&&B[i].y>R[j].y&&R[j].used==0)
			{
				if(B[i].y-R[j].y<miny)
				     {
				     	miny=(B[i].y-R[j].y);
				     	y=j;
					 }
			}
			
		}
		R[y].used=1;
		if(y!=-1) cnt++;
	}
	cout<<cnt<<endl;
}

### 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;
struct point{
	int x;
	int y;
	int used;
}R[105],B[105];
bool cmp(point a,point b)
{
    return a.x<b.x;
}
int main()
{
	int n;
	int a,b;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d %d",&R[i].x,&R[i].y);
	for(int i=0;i<n;i++)
		scanf("%d %d",&B[i].x,&B[i].y); 
	sort(B,B+n,cmp);
	int cnt=0;
	for(int i=0;i<n;i++)
	{
		int miny=200;
		int y=-1;
		for(int j=0;j<n;j++)
		{
			if(B[i].x>R[j].x&&B[i].y>R[j].y&&R[j].used==0)
			{
				if(B[i].y-R[j].y<miny)
				     {
				     	miny=(B[i].y-R[j].y);
				     	y=j;
					 }
			}
			
		}
		R[y].used=1;
		if(y!=-1) cnt++;
	}
	cout<<cnt<<endl;
}

```