#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
bool f[110];
int main(){
  ll n;
  cin>>n;
  P a[110],b[110];
  for(int i=0;i<n;i++)cin>>a[i].first>>a[i].second;
  for(int i=0;i<n;i++)cin>>b[i].first>>b[i].second;
  sort(a,a+n);
  sort(b,b+n);
  ll ans=0;
  for(int i=0;i<n;i++){
    P num,yan;
    num=make_pair(-1,-1);
    yan=num;
    for(int j=0;j<n;j++){
      if(f[j])continue;
      if(a[j].first<b[i].first&&a[j].second<b[i].second){
        if(num.second<a[j].second){
          num.second=a[j].second;
          num.first=j;
        }
      }
    }
    if(num!=yan){
      ans++;
      f[num.first]=1;
    }
  }
  cout<<ans<<endl;
  return 0;
}
### 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;
typedef long long ll;
typedef pair<ll,ll> P;
bool f[110];
int main(){
  ll n;
  cin>>n;
  P a[110],b[110];
  for(int i=0;i<n;i++)cin>>a[i].first>>a[i].second;
  for(int i=0;i<n;i++)cin>>b[i].first>>b[i].second;
  sort(a,a+n);
  sort(b,b+n);
  ll ans=0;
  for(int i=0;i<n;i++){
    P num,yan;
    num=make_pair(-1,-1);
    yan=num;
    for(int j=0;j<n;j++){
      if(f[j])continue;
      if(a[j].first<b[i].first&&a[j].second<b[i].second){
        if(num.second<a[j].second){
          num.second=a[j].second;
          num.first=j;
        }
      }
    }
    if(num!=yan){
      ans++;
      f[num.first]=1;
    }
  }
  cout<<ans<<endl;
  return 0;
}
```