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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define rep(i, a) for (int i = (int)0; i < (int)a; ++i)
#define rrep(i, a) for (int i = (int)a - 1; i >= 0; --i)
#define REP(i, a, b) for (int i = (int)a; i < (int)b; ++i)
#define RREP(i, a, b) for (int i = (int)a - 1; i >= b; --i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define popcount __builtin_popcount
using ll = long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;

template <class T>
inline bool chmin(T &a, T b)
{
	if (a > b)
	{
		a = b;
		return true;
	}
	return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
	if (a < b)
	{
		a = b;
		return true;
	}
	return false;
}

ll gcd(ll n, ll m)
{
	ll tmp;
	while (m != 0)
	{
		tmp = n % m;
		n = m;
		m = tmp;
	}
	return n;
}

ll lcm(ll n, ll m)
{
	return abs(n) / gcd(n, m) * abs(m); //gl=xy
}

using namespace std;

using P=pair<int,int>;
void solve()
{
	int n;
	cin>>n;
	vector<P>red(n),blue(n);
	rep(i,n)cin>>red[i].second>>red[i].first;
	rep(i,n)cin>>blue[i].first>>blue[i].second;
	sort(rall(red));sort(rall(blue));
	rep(i,n)swap(red[i].first,red[i].second);
	vector<bool>used(n);
	int ans=0;
	rep(i,n){
		rep(j,n){
			if(used[j])continue;
			if(red[j].first<blue[i].first&&red[j].second<blue[i].second){
				used[j]=true;
				++ans;
				break;
			}
		}
	}
	cout<<ans<<"\n";
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(15);
	solve();
	return 0;
}