#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int,int>;
#define rep(i,s,n) for(int i = s; i < (int)(n); i++)

int main() {
  int n,ans = 0;
  cin >> n;
  vector<P>a(n),b(n);
  rep(i,0,n) cin >> a[i].first >> a[i].second;
  rep(i,0,n) cin >> b[i].first >> b[i].second;
  sort(b.begin(),b.end());
  vector<bool>used(n,false);
  rep(i,0,n){
    int x = b[i].first,y = b[i].second,my = -1,py;
    rep(j,0,n){
      if(a[j].first < x && a[j].second > my && a[j].second < y && !used[j]){
        my = a[j].second;
        py = j;
      }
    }
    if(my != -1){
      ans++;
      used[py] = true;
    }
  }
  cout << ans << endl;
}
### Prompt

Generate 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;
using P = pair<int,int>;
#define rep(i,s,n) for(int i = s; i < (int)(n); i++)

int main() {
  int n,ans = 0;
  cin >> n;
  vector<P>a(n),b(n);
  rep(i,0,n) cin >> a[i].first >> a[i].second;
  rep(i,0,n) cin >> b[i].first >> b[i].second;
  sort(b.begin(),b.end());
  vector<bool>used(n,false);
  rep(i,0,n){
    int x = b[i].first,y = b[i].second,my = -1,py;
    rep(j,0,n){
      if(a[j].first < x && a[j].second > my && a[j].second < y && !used[j]){
        my = a[j].second;
        py = j;
      }
    }
    if(my != -1){
      ans++;
      used[py] = true;
    }
  }
  cout << ans << endl;
}
```