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 <iostream>
#include <string>
#include <vector>
#include <algorithm> // sort
#include <map> // pair
using namespace std;

int main(){
  int count = 0;
  int N;
  cin >> N;
  vector<pair<int, int> > red(N);
  vector<pair<int, int> > usedred(N);
  for(int i=0; i<N; i++){
    cin >> red[i].first >> red[i].second;
    usedred[i].first = -1;
    usedred[i].second = -1;
  }
  vector<pair<int, int> > blue(N);
  vector<pair<int, int> > usedblue(N);
  for(int i=0; i<N; i++){
    cin >> blue[i].first >> blue[i].second;
    usedblue[i].first = -1;
    usedblue[i].second = -1;
  }
  sort(red.begin(), red.end());
  sort(blue.begin(), blue.end());

  int ii = 0;
  for(int i=0; i<N; i++){
    for(int j=0; j<N; j++){
      if(usedred[i].first!=0 && usedred[i].second!=0 && usedblue[j].first!=0 && usedblue[j].second!=0){
        if(red[i].first < blue[j].first){
          if(red[i].second < blue[j].second){
            ii=i;
            while(red[ii].second < blue[j].second){
                ii++;
            }
            usedred[ii-1].first = 0;
            usedred[ii-1].second = 0;
            usedblue[j].first = 0;
            usedblue[j].second = 0;
            count++;
          }
        }
      }

    }
  }
  cout << count << endl;
return 0;
}