#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define SX(x) (x).begin(), (x).end()
typedef pair<int, int> P;

int main()
{
  int n;	cin >> n;
  vector<P> a(n), b(n);
  rep(i,n) cin >> a[i].first >> a[i].second;
  rep(i,n) cin >> b[i].first >> b[i].second;
  sort(SX(a));
  sort(SX(b));
  
  int ans=0;
  rep(i,n) {
    int res=-1, tmp =-1;
    rep(j, a.size())
    {
      if (a[j].first < b[i].first && a[j].second < b[i].second)
      {
        if (a[j].second > res)
        {
          tmp = j;
          res = a[j].second;
        }
      }
    }
    if (res != -1)
    {
      a.erase(a.begin() + tmp);
      ans++;
    }
  }
  cout << ans << endl;
}
### 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 rep(i,n) for(int i=0; i<(int)(n); i++)
#define SX(x) (x).begin(), (x).end()
typedef pair<int, int> P;

int main()
{
  int n;	cin >> n;
  vector<P> a(n), b(n);
  rep(i,n) cin >> a[i].first >> a[i].second;
  rep(i,n) cin >> b[i].first >> b[i].second;
  sort(SX(a));
  sort(SX(b));
  
  int ans=0;
  rep(i,n) {
    int res=-1, tmp =-1;
    rep(j, a.size())
    {
      if (a[j].first < b[i].first && a[j].second < b[i].second)
      {
        if (a[j].second > res)
        {
          tmp = j;
          res = a[j].second;
        }
      }
    }
    if (res != -1)
    {
      a.erase(a.begin() + tmp);
      ans++;
    }
  }
  cout << ans << endl;
}
```