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;

#define rep(i, n) REP(i, 0, n)
#define REP(i,k,n) for(int i = k; i < n; i++)
#define mmax(start, end) *max_element(start, end)
#define mmin(start, end) *min_element(start, end)
#define argmax max_element
#define argmin min_element
#define push push_back
#define pop pop_back
#define ALL(x) (x).begin(), (x).end()

typedef long long ll;
typedef string str;

int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};

template<class T = ll> T in() {
    T _in;
    cin >> _in;
    return _in;
}

// ソート用
bool comp(const pair<int, int>& x, const pair<int, int>& y) {
    if(x.first == y.first)
        return x.second < y.second;
    else
        return x.first < y.first;
}

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

    int N;
    cin >> N;
    pair<int, int> blue[100], red[100];

    rep(i, N) {
        red[i] = make_pair(in<int>(), in<int>());
    }

    rep(i, N) {
        blue[i] = make_pair(in<int>(), in<int>());
    }

    sort(blue, blue+N, comp);
    sort(red, red+N, comp);

    // rep(i, N) {
    //     cout << red[i].first << " " << red[i].second << " ";
    // }
    // cout << endl << endl;
    //
    // rep(i, N) {
    //     cout << blue[i].first << " " <<  blue[i].second << " ";
    // }
    // cout << endl << endl;

    // 数える
    int ans = 0;
    rep(i, N) {
        rep(j, N) {
            if(red[i].first<blue[j].first && red[i].second<blue[j].second) {
                // ペア成立，当該青点を使用不可にする
                blue[j].first = blue[j].second = -1;
                // cout << i << " " << j << endl;
                ans++;
                break;
            }
            if(j == N-1) {
                // これ以上はペアを作れない
                cout << ans << endl;
                return 0;
            }
        }
    }

    cout << ans << endl;
    return 0;
}