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

bool f_sort(pair<int,int> a,pair<int,int> b){return a.first<b.first;};
bool s_sort(pair<int,int> a,pair<int,int> b){return a.second<b.second;};
int main(){
	int n; cin>>n;
	int cnt=0;
	pair<int,int> red[n],blue[n];
	for(int i=0;i<n;i++) cin>>red[i].first>>red[i].second;
	for(int i=0;i<n;i++) cin>>blue[i].first>>blue[i].second;
	sort(red,red+n,s_sort);
	sort(blue,blue+n,f_sort);	for(int i=0;i<n;i++){
		for(int j=n-1;j>=0;j--){
			if(red[j].first<blue[i].first && red[j].second<blue[i].second && red[j].first!=-1){
				cnt++;
				red[j].first=-1;
				break;
			}
		}
	}
	cout<<cnt<<endl;
}

### Prompt

In CPP, your task is to solve the following problem:

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;

bool f_sort(pair<int,int> a,pair<int,int> b){return a.first<b.first;};
bool s_sort(pair<int,int> a,pair<int,int> b){return a.second<b.second;};
int main(){
	int n; cin>>n;
	int cnt=0;
	pair<int,int> red[n],blue[n];
	for(int i=0;i<n;i++) cin>>red[i].first>>red[i].second;
	for(int i=0;i<n;i++) cin>>blue[i].first>>blue[i].second;
	sort(red,red+n,s_sort);
	sort(blue,blue+n,f_sort);	for(int i=0;i<n;i++){
		for(int j=n-1;j>=0;j--){
			if(red[j].first<blue[i].first && red[j].second<blue[i].second && red[j].first!=-1){
				cnt++;
				red[j].first=-1;
				break;
			}
		}
	}
	cout<<cnt<<endl;
}

```