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: Runtime Error)

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 1100;

struct Edge{
    int to, next;
};

Edge Edges[MAXN << 1];
int head[MAXN], cnt, Nx, Ny;

int Mx[MAXN], My[MAXN];
queue<int> Q;
int dx[MAXN], dy[MAXN];
bool vis[MAXN];

bool HopcroftKarpDfs(int u){
    for(int i = head[u]; ~i; i = Edges[i].next){
        int v = Edges[i].to;
        if(!vis[v] && dy[v] == dx[u] + 1){
            vis[v] = true;
            if(!My[v] || HopcroftKarpDfs(My[v])){
                Mx[u] = v, My[v] = u;
                return true;
            }
        }
    }
    return false;
}

int HopcroftKarp(){
    memset(Mx, 0, sizeof(Mx));
    memset(My, 0, sizeof(My));
    int ans = 0;
    while(true){
        bool flag = false;
        while(!Q.empty()) Q.pop();
        memset(dx, 0, sizeof(dx));
        memset(dy, 0, sizeof(dy));
        for(int i = 1; i <= Nx; i++) if(!Mx[i]) Q.push(i);
        while(!Q.empty()){
            int u = Q.front();
            Q.pop();
            for(int i = head[u]; ~i; i = Edges[i].next){
                int v = Edges[i].to;
                if(!dy[v]){
                    dy[v] = dx[u] + 1;
                    if(My[v]){
                        dx[My[v]] = dy[v] + 1;
                        Q.push(My[v]);
                    }else flag = true;
                }
            }
        }
        if(!flag) break;
        memset(vis, false, sizeof(vis));
        for(int i = 1; i <= Nx; i++) if(!Mx[i] && HopcroftKarpDfs(i)) ans++;
    }
    return ans;
}

int a[MAXN], b[MAXN], c[MAXN], d[MAXN];

int main()
{
    int N;

    cin >> N;

    for (int i = 1; i <= N; i++) {
        cin >> a[i] >> b[i];
    }

    for (int i = 1; i <= N; i++) {
        cin >> c[i] >> d[i];
    }

    memset(head, -1, sizeof(head));
    cnt = 0;

    for (int i = 1; i <= N; i++) {
        for (int j = N + 1; j <= (N << 1); j++) {
            if (a[i] < c[j - N] && b[i] < d[j - N]) {
                Edges[cnt].to = i, Edges[cnt].next = head[j], head[j] = cnt++;
                Edges[cnt].to = j, Edges[cnt].next = head[i], head[i] = cnt++;
            }
        }
    }

    Nx = Ny = (N << 1);
    int ans = HopcroftKarp();

    cout << ans / 2;

    return 0;
}