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 <bits/stdc++.h>
using namespace std;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;

#define MAX_V 200

int V;//頂点数
vector<int> G[MAX_V];//グラフの隣接リスト表現
int match[MAX_V];//マッチングのペア
bool used[MAX_V];//DFSですでに調べたかのフラグ

//uとvを結ぶ辺をグラフに追加する
void add_edge(int u, int v){
    G[u].push_back(v);
    G[v].push_back(u);
}

//増加パスをDFSで探す
bool dfs(int v){
    used[v] = true;
    for(int i = 0; i < G[v].size(); i++){
        int u = G[v][i], w = match[u];
        if(w < 0 || !used[w] && dfs(w)){
            match[v] = u;
            match[u] = v;
            return true;
        }
    }
    return false;
}

//二部グラフの最大マッチングを求める
int bipartite_matching(){
    int res = 0;
    memset(match, -1, sizeof(match));
    for(int v = 0; v < V; v++){
        if(match[v] < 0){
            memset(used, 0, sizeof(used));
            if(dfs(v)){
                res++;
            }
        }
    }
    return res;
}

signed main(){
    int n;
    cin >> n;
    int a[n],b[n];
    REP(i,n){
        cin >> a[i] >> b[i];
    }
    int c[n],d[n];
    REP(i,n){
        cin >> c[i] >> d[i];
    }
    REP(i,n){
        REP(j,n){
            if(a[i] < c[j] && b[i] < d[j]){
                add_edge(i, n + j);
            }
        }
    }
    V = 2 * n;
    cout << bipartite_matching() << endl;
}