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 < pair<int,int> > v,d;
    for(int i=0;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        v.push_back(make_pair(x,y));
    }
    
    for(int i=0;i<n;i++)
    {
        int x,y;
        cin>>x>>y;
        d.push_back(make_pair(x,y));
    }int sum=0;int y[10000]={};
    sort(v.begin(),v.end());
    sort(d.begin(),d.end());
    for(int i=0;i<n;i++)
    {int j=n-1;int o=j;
        while(v[j].first>d[i].first||v[j].second>d[i].second&&j>=0||y[j]==-1)
        {j--;
            if(y[j]!=-1)
            o=j;
        }
        if(v[o].first<d[i].first&&v[o].second<d[i].second&&o>=0&&y[o]!=-1)
        {sum++;y[o]=-1;}
    }
    cout<<sum;
    return 0;
}