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<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define pn(s) cout << (#s) << " " << (s) << endl

const ll mod = 1e9 + 7;
const ll inf = 1e18;

class Point{
    public:
    Point(int a, int b){
        x = a;
        y = b;
    }
    int x;
    int y;

    bool operator<(const Point &another) const
    {
        return x < another.x;
    };

    bool operator==(const Point &another) const{
        return x==another.x && y==another.y;
    }

    static bool sortByY(Point a, Point b){
        if(a.y == b.y){
            return a.x < a.y;
        }else{
            return a.y < b.y;
        }
    }
};

void PrintPointVector(vector<Point> P){
    p("[vector]");
    if(P.size()==0){
        p("nothing");
    }
    for(auto p : P){
        cout << p.x << " " << p.y << endl;
    }
}


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    int N;
    cin >> N;

    vector<Point> R;
    FOR(i, 0, N){
        int x, y;
        cin >> x >> y;
        R.push_back(Point(x, y));
    }

    vector<Point> B;
    FOR(i, 0, N){
        int x, y;
        cin >> x >> y;
        B.push_back(Point(x, y));
    }

    // Bをxの小さい順に並べる
    sort(ALL(B));

    int pairNum = 0;

    for(auto blue : B){
        // blue.xよりxが小さいものだけ集める
        vector<Point> temp;
        for(auto red : R){
            if(red.x < blue.x && red.y < blue.y){
                temp.push_back(red);
            }
        }

        if(temp.size() > 0){
            // yの小さい方からソート
            sort(ALL(temp), Point::sortByY);

            // yが最大の点をペアに選ぶ
            auto selectedRed = temp[temp.size()-1];

            // それをRから消す
            auto it = find(ALL(R), selectedRed);
            R.erase(it);
            pairNum++;
        }
    }

    p(pairNum);
    
    return 0;
}