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: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<utility>
#include<cmath>
#include<climits>
#include<queue>
#include<stack>
#include<numeric>
#include<set>
#include<iomanip>
using namespace std;


typedef long long ll;
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)



int main(){
  int n;
  cin >> n;
  vector<pair<int, int> > red(n), blue(n);
  rep(i, n) cin >> red[i].first >> blue[i].second;
  rep(i, n) cin >> blue[i].first >> blue[i].second;
  vector<bool> user(n, true), useb(n, true);

  int ans = 0;
  sort(red.begin(), red.end());
  sort(blue.begin(), blue.end());
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < n; ++j){
      if(red[i].first < blue[j].first && red[i].second < blue[j].second && user[i] && useb[j]){
        int tmp = j;
        for(int k = j; k < n; ++k){
          if(red[i].first < blue[j].first && red[i].second < blue[j].second){
            if(red[i].second > red[tmp].second){
              tmp = i;
            }
          }
        }
        user[i] = false;
        useb[tmp] = false;
        ans++;
      }
    }
  }

  cout << ans << endl;


}