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

int main()
{
    int n;cin>>n;
    vector<int> a(n);
    vector<int> b(n);
    vector<int> c(n);
    vector<int> d(n);
    int i,j;
    for(i=0;i<n;i++)cin>>a.at(i)>>b.at(i);
    for(i=0;i<n;i++)cin>>c.at(i)>>d.at(i);
    vector<pair<int,int>> p1(n);
    vector<pair<int,int>> p2(n);
    vector<bool> taken1(n,false);
    vector<bool> taken2(n,false);
    for(i=0;i<n;i++)
    {
        p1.at(i)=make_pair(a.at(i),b.at(i));
        p2.at(i)=make_pair(c.at(i),d.at(i));
    }
    sort(p1.begin(),p1.end());
    reverse(p1.begin(),p1.end());
    sort(p2.begin(),p2.end());
    reverse(p2.begin(),p2.end());
    int ans=0;
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(p1.at(i).first<p2.at(j).first&&p1.at(i).second<p2.at(j).second&&!taken1.at(i)&&!taken2.at(j))
            {
                taken1.at(i)=true;
                taken2.at(j)=true;
                ans++;
                break;
            }
        }
    }
    cout<<ans<<endl;
}