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: Accepted)

#include <iostream>
#include <string>
#include <unordered_map>
#include <stdio.h>
#include <list>
#include <vector>
#include <algorithm>
#include <stack>
#include <climits>
#include <cmath>
#include <map>
#include <queue>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; ++i)
#define ALL(v) v.begin(), v.end()
typedef long long ll;

int main()
{
	int n ;cin >> n;
	vector<bool> ispairAB(n);
	vector<bool> ispairCD(n);
	rep(i,n)
	{
		ispairAB[i]=false;
		ispairCD[i]=false;
	}
	vector<pair<int, int>> AB;
	vector<pair<int, int>> CD;
	rep(i,n)
	{
		int foo,fooo;
		cin >> foo >> fooo;
		AB.emplace_back(foo,fooo);
	}
	rep(i,n)
	{
		int foo,fooo;
		cin >> foo >> fooo;
		CD.emplace_back(foo,fooo);
	}
	sort(ALL(AB));
	sort(ALL(CD));
	int c=0;
	rep(i,n)
	{
		int max=-1;
		int maxnum=-1;
		for(int j=0;j<n&&AB[j].first<CD[i].first;j++)
		{
			if( (!ispairAB[j])&&(!ispairCD[i]))
			if(AB[j].second<CD[i].second)
			{	
				if(max<AB[j].second)
				{
					max=AB[j].second;
					maxnum = j;
				}
			}
		}
		if(max!=-1)
		{
			ispairCD[i]=true;
			ispairAB[maxnum]=true;
			c++;
		}
	}
	cout<<c<<endl;
	return 0;
}