Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
#define fi first
#define se second
int main(){
    int n;
    cin>>n;
    vector<P> a(n),b(n);
    for(int i=0;i<n;i++){
       int x,y;
       cin>>x>>y;
       a.at(i)=P(x,y);
    }
    for(int i=0;i<n;i++){
       int x,y;
       cin>>x>>y;
       b.at(i)=P(x,y);
    }
    sort(a.begin(),a.end(),greater<P>());
    vector<bool> b1(n,true);
    int count=0;
  for(int i=0;i<n;i++){
      priority_queue<P> u;
      for(int j=0;j<n;j++){
          if(b1.at(j)&&a.at(i).fi<b.at(j).fi&&a.at(i).se<b.at(j).se)u.push({b.at(j).se,b.at(j).fi});
      }
      if(u.size()==0)continue;
      count++;
      P c=u.top();
      P d={c.fi,c.se};
      int check;
      for(int k=0;k<n;k++){
          if(b.at(k)==d){
              check=k;
              break;
          }   
      }
     b1.at(check)=false;
  }
cout<<count<<endl;
}