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;
typedef long long ll;
typedef pair<ll, ll> P;
#define vec vector
#define fi first
#define se second
#define rep(i,n) for (ll i = 0; i < (n); ++i)
#define vcout(v) rep(i,v.size()) cout << v[i] << " "; cout << endl
#define mcout(m) rep(j,m.size()){vcout(m[j]);}
#define bit(x) (1ll<<(x))

void mainsolve(){
	
	ll n;
	cin >> n;
	vec<P> red(n), blue(n);
	rep(i,n){
		ll a,b;
		cin >> a >> b;
		red[i] = P(a,b);
	}
	rep(i,n){
		ll c,d;
		cin >> c >> d;
		blue[i] = P(c,d);
	}
	
	vec<vec<ll>> redto(n); // red[i]はblue[j]とペアになれる
	vec<vec<ll>> blueto(n); // blue[i]はred[j]とペアになれる
	rep(i,n){
		rep(j,n){
			P rr = red[i];
			P bb = blue[j];
			if(rr.fi < bb.fi && rr.se < bb.se){
				redto[i].push_back(j);
				blueto[j].push_back(i);
			}
		}
	}
	
	vec<ll> cnt(n); // red[i]の相手候補数
	rep(i,n) cnt[i] = redto[i].size();
	
	ll res = 0;
	vec<bool> checked(n,false);
	rep(i,n){
		ll aite = -1;
		ll tmp = 1e9;
		for(auto rr : blueto[i]){
			if(!checked[rr] && tmp > cnt[rr]){
				aite = rr;
				tmp = cnt[rr];
			}
		}
		if(aite!=-1){
			checked[aite] = true; // もう選ばれない
			res++;
		}
	}
	
	cout << res << endl;
	
}

int main(){
	
	//ll TC;cin>>TC;rep(I,TC){cout<<"case:"<<I+1<<"---"<<endl; // comment out!
		mainsolve();
	//} // comment out!
	return 0;
}