#include <iostream>
#include <vector>
#include <utility>
#include <set>
using namespace std;

vector<pair<int, bool>> v(210);
int main() {
	int N; cin >> N;
	for(int i=0; i<N*2; ++i){
	    int a, b;
	    cin >> a >> b;
	    v[a]=make_pair(b, i<N);
	}
	set<int> st;
	int ans=0;
	for(int i=0; i<N*2; ++i){
	    if(v[i].second) st.insert(v[i].first);
	    else{
	        for(int j=v[i].first-1; j>=0; --j){
	            if(st.find(j)!=st.end()){
	                ++ans;
	                st.erase(j);
	                break;
	            }
	        }
	    }
	}
	cout << ans << endl;
	return 0;
}

### Prompt

Create a solution in Cpp for 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 <iostream>
#include <vector>
#include <utility>
#include <set>
using namespace std;

vector<pair<int, bool>> v(210);
int main() {
	int N; cin >> N;
	for(int i=0; i<N*2; ++i){
	    int a, b;
	    cin >> a >> b;
	    v[a]=make_pair(b, i<N);
	}
	set<int> st;
	int ans=0;
	for(int i=0; i<N*2; ++i){
	    if(v[i].second) st.insert(v[i].first);
	    else{
	        for(int j=v[i].first-1; j>=0; --j){
	            if(st.find(j)!=st.end()){
	                ++ans;
	                st.erase(j);
	                break;
	            }
	        }
	    }
	}
	cout << ans << endl;
	return 0;
}

```