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 <vector>
#include <string>
#include <algorithm>

using namespace std;

#define ll long long
#define vi vector<int>
#define pii pair<int, int>
#define vpii vector< pair<int, int> >
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

int N, cnt = 0;
bool isUsed[105]={};
vpii Red, Blue;

void print_Red(){
    cout << "Red" << endl;
    rep(i,N)
        printf("(%d,%d)\n", Red[i].first, Red[i].second);
}

void print_Blue(){
    cout << "Blue" << endl;
    rep(i,N)
        printf("(%d,%d)\n", Blue[i].first, Blue[i].second);
}

bool isOk(pii r, pii b){
    return r.first < b.first && r.second < b.second;
}

int calc_dist(pii r, pii b){
    return pow(b.first-r.first,2) + pow(b.second-r.second,2);
}

void solve(){
    for(int i = 0; i < N; i++){
        int min_id = -1, max_y = -1;
        for(int j = 0; j < N; j++){
            if(isOk(Red[j], Blue[i]) && !isUsed[j]){
                    int buff_y = Red[j].second;
                    if(buff_y >  max_y){
                        min_id = j;
                        max_y = buff_y;
                    }
                }
            }
        if(min_id != -1) isUsed[min_id] = true;
        }

    rep(i,N)
        if(isUsed[i]) cnt++;

    cout << cnt << endl;
}

int main(){
    cin >> N;
    int buff1, buff2;

    rep(i,N){
        cin >> buff1 >> buff2;
        Red.push_back(make_pair(buff1, buff2));
    }
    rep(i,N){
        cin >> buff1 >> buff2;
        Blue.push_back(make_pair(buff1, buff2));
    } 

    sort(Red.begin(), Red.end());
    sort(Blue.begin(), Blue.end());
    // print_Red();
    // print_Blue();

    solve();
}