#include <bits/stdc++.h>
using namespace std;
#define P pair <int,int>

int N;
const int MAXN = 105;
P A[MAXN],B[MAXN];
bool f[MAXN];
int ans;
bool mycmp(P a,P b){
	return a.first<b.first;
}

int main(){
	scanf("%d",&N);
	for(int i=0;i<N;i++) scanf("%d%d",&A[i].first,&A[i].second);	
	for(int i=0;i<N;i++) scanf("%d%d",&B[i].first,&B[i].second);

	sort(A,A+N,mycmp);
	sort(B,B+N,mycmp);

	for(int i=0;i<N;i++){
		int maxx = 0;int t = 1000;
		for(int j=0;j<N;j++){
			if(A[j].first>B[i].first) break;
			if(f[j]==0 & A[j].second<B[i].second){
				if(A[j].second>=maxx){
				 	maxx = A[j].second;
					t=j;
				}	
			}
			
		}
		if(t!=1000){
			f[t]=1;
			ans++;
		}
	}

	printf("%d\n",ans);
}
### 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;
#define P pair <int,int>

int N;
const int MAXN = 105;
P A[MAXN],B[MAXN];
bool f[MAXN];
int ans;
bool mycmp(P a,P b){
	return a.first<b.first;
}

int main(){
	scanf("%d",&N);
	for(int i=0;i<N;i++) scanf("%d%d",&A[i].first,&A[i].second);	
	for(int i=0;i<N;i++) scanf("%d%d",&B[i].first,&B[i].second);

	sort(A,A+N,mycmp);
	sort(B,B+N,mycmp);

	for(int i=0;i<N;i++){
		int maxx = 0;int t = 1000;
		for(int j=0;j<N;j++){
			if(A[j].first>B[i].first) break;
			if(f[j]==0 & A[j].second<B[i].second){
				if(A[j].second>=maxx){
				 	maxx = A[j].second;
					t=j;
				}	
			}
			
		}
		if(t!=1000){
			f[t]=1;
			ans++;
		}
	}

	printf("%d\n",ans);
}
```