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<vector>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<numeric>
#include<iomanip>
#include<deque>
#include<tuple>
#include<queue>
#include<map>
#include<cstdint>
#include<unordered_map>
#include<boost/multiprecision/cpp_int.hpp>


#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define F first
#define S second

namespace mp = boost::multiprecision;

using cpp_int = mp::cpp_int;
using ll = long long;
using namespace std;
typedef pair<int, int> P;

bool pairsort1(const P& element1, const P& element2)
{//first昇順
	return element1.F < element2.F;
}
bool pairsort2(const P& element1, const P& element2)
{//first降順
	return element1.F > element2.F;
}
bool pairsort3(const P& element1, const P& element2)
{//second昇順
	return element1.S < element2.S;
}
bool pairsort4(const P& element1, const P& element2)
{//second降順
	return element1.S > element2.S;
}

int main() {
	int N;
	cin >> N;
	vector <pair<int, int>>ab(N), cd(N);
	vector<vector<bool>>map(N*2, vector<bool>(N*2,false));
	rep(i, N) {
		cin >> ab[i].F >> ab[i].S;
		map[ab[i].F][ab[i].S]=true;
	}
	rep(i, N) {
		cin >> cd[i].F >> cd[i].S;
	}
	sort(all(cd), pairsort1);
	sort(all(ab),pairsort1);
	vector<pair<int,int>>pair(N);
	int atmp = 0 , btmp = 0;
	int count = 0;
	rep(i, N) {
		int j = 0;
		int atmp = 0;
		int btmp = -1;
		bool flag = false;
		while (ab[j].F < cd[i].F){
			if (ab[j].S<cd[i].S&&
				ab[j].S >btmp&&
				map[ab[j].F][ab[j].S] ==true) {
				atmp = ab[j].F;
				btmp = ab[j].S;
				flag = true;
			}
			j++;
			if (j == N) {
				break;
			}
		}
		if (flag) {
			count++;
			map[atmp][btmp] = false;
		}
	}
	cout << count << endl;
}