#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const ll mod=1e9+7;
int dy[]={1,-1,0,0};
int dx[]={0,0,1,-1};


int main(){
  int N;
  cin>>N;
  vector<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.rbegin(),red.rend());

  int ans=0;
  vector<bool> used(N,false);
  for(int i=0;i<N;i++){
    int place=-1;
    int miny=1000;
    for(int j=0;j<N;j++){
      if(!used[j]&&red[i].first<blue[j].first&&red[i].second<blue[j].second&&blue[j].second<miny){
        place=j;
        miny=blue[j].second;
      }
    }
    if(place==-1) continue;
    ans++;
    used[place]=true;

  }
  cout<<ans<<endl;
  return 0;
}


### Prompt

Your challenge is to write 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;
using ll=long long;
const ll mod=1e9+7;
int dy[]={1,-1,0,0};
int dx[]={0,0,1,-1};


int main(){
  int N;
  cin>>N;
  vector<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.rbegin(),red.rend());

  int ans=0;
  vector<bool> used(N,false);
  for(int i=0;i<N;i++){
    int place=-1;
    int miny=1000;
    for(int j=0;j<N;j++){
      if(!used[j]&&red[i].first<blue[j].first&&red[i].second<blue[j].second&&blue[j].second<miny){
        place=j;
        miny=blue[j].second;
      }
    }
    if(place==-1) continue;
    ans++;
    used[place]=true;

  }
  cout<<ans<<endl;
  return 0;
}


```