#include <bits/stdc++.h>
using namespace std;
#define modulo 998244353
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000000



int main(){
	
	int N;
	cin>>N;
	
	vector<pair<pair<int,int>,char>> P(2*N);
	
	for(int i=0;i<2*N;i++){
		int x,y;
		cin>>x>>y;
		P[i].first = make_pair(x,y);
		if(i<N)P[i].second = 'R';
		else P[i].second = 'B';
	}
	
	sort(P.begin(),P.end());
	
	int ans = 0;
	set<int> y;
	
	for(int i=0;i<2*N;i++){
		if(P[i].second=='R')y.insert(P[i].first.second);
		else{
			auto it = y.lower_bound(P[i].first.second);
			if(it==y.begin())continue;
			it--;
			y.erase(it);
			ans++;
		}
	}
	
	cout<<ans<<endl;
	

		
    return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
#define modulo 998244353
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 10000000000000000



int main(){
	
	int N;
	cin>>N;
	
	vector<pair<pair<int,int>,char>> P(2*N);
	
	for(int i=0;i<2*N;i++){
		int x,y;
		cin>>x>>y;
		P[i].first = make_pair(x,y);
		if(i<N)P[i].second = 'R';
		else P[i].second = 'B';
	}
	
	sort(P.begin(),P.end());
	
	int ans = 0;
	set<int> y;
	
	for(int i=0;i<2*N;i++){
		if(P[i].second=='R')y.insert(P[i].first.second);
		else{
			auto it = y.lower_bound(P[i].first.second);
			if(it==y.begin())continue;
			it--;
			y.erase(it);
			ans++;
		}
	}
	
	cout<<ans<<endl;
	

		
    return 0;
}

```