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 <cmath>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <numeric>
#include <functional>
#include <map>
#define REP(i, n) for(int i=0;i<n;++i)
#define INF 100000000
#define ll long long

using namespace std;

typedef pair<int, int> P;

int n;
int a, b, c, d;
int ans=0;
vector<P> v, w;;

int main(){
  cin >> n;
  REP(i, n){
    cin >> a >> b;
    v.push_back(P(a, b));
  }
  REP(i, n){
    cin >> c >> d;
    w.push_back(P(c, d));
  }

  sort(w.begin(), w.end());

  REP(i, n){
    int s,t, itr;
    s = w[i].first;
    t = w[i].second;

    int max=-1;
    bool b = false;
    REP(j, v.size()){
      if((v[j].first<s) && (v[j].second<t) && (v[j].second>max) ){
        itr = j;
        max = v[j].second;
        b = true;
      }
    }

    if(b){
      ans += 1;
      v.erase(v.begin() + itr);
    }
  }

  cout << ans << endl;

  return 0;
}