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 <algorithm>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <functional>
#include <iostream>
#include <map>
#include <memory>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
#include <iomanip>
using namespace std;

typedef long long ll;

#define sz size()
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(c) (c).begin(), (c).end()
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define per(i,a,b) for(ll i=(b-1);i>=(a);--i)
#define clr(a, b) memset((a), (b) ,sizeof(a))
#define ctos(c) string(1,c)
#define print(x) cout<<#x<<" = "<<x<<endl;

#define MOD 1000000007

#define PMAX_A 110
#define PMAX_B 110

#define NN (PMAX_A+PMAX_B+2)

int vv[NN][NN];
int flag[NN];

int flow(int s, int e, int f){
  if(s == e)return f;

  flag[s] = 1;

  for(int i = 0; i < NN; i++){
    if(vv[s][i] > 0 && flag[i] == 0){
      int d = flow(i, e, min(f, vv[s][i]));
      if(d > 0){
        vv[s][i] -= d;
        vv[i][s] += d;
        return d;
      }
    }
  }
  return 0;
}

void ae(int from, int to, int cost){
  vv[from][to] = cost;
  vv[to][from] = 0;
}

int pmatch(int pa[PMAX_A][PMAX_B]){
  memset(vv,-1,sizeof(vv));

  for(int i = 1; i <= PMAX_A; i++){
    ae(0,i,1);
  }

  for(int i = 0; i < PMAX_A; i++){
    for(int j = 0; j < PMAX_B; j++){
      if(pa[i][j] == 1){
        ae(i+1,j+PMAX_A+1,1);
      }
    }
  }

  for(int i = PMAX_A+1; i <= PMAX_A+PMAX_B; i++){
    ae(i,PMAX_A+PMAX_B+1,1);
  }

  int ans = 0;

  for(;;){
    memset(flag,0,sizeof(flag));
    int f = flow(0, PMAX_A+PMAX_B+1, 1000000000);
    if(f == 0)break;
    ans += f;
  }

  return ans;
}

int main() {
  int pa[PMAX_A][PMAX_B];

  memset(pa,0,sizeof(pa));

  ll n;
  cin>>n;
  vector<ll> va1;
  vector<ll> va2;
  rep(i,0,n){
  	ll a,b;
  	cin>>a>>b;
  	va1.pb(a);
  	va2.pb(b);
  }
  vector<ll> vb1;
  vector<ll> vb2;
  rep(i,0,n){
  	ll a,b;
  	cin>>a>>b;
  	vb1.pb(a);
  	vb2.pb(b);
  }
  rep(i,0,va1.sz){
  	rep(j,0,vb1.sz){
  		if(va1[i]<vb1[j]&&va2[i]<vb2[j]){
  			pa[i][j] = 1;
  		}
  	}
  }
  cout << pmatch(pa) << endl;
	return 0;
}