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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define inf (int)(1e9 + 7)

typedef struct {
	int *array;
	int last;
	int first;
}queue;

queue *make_queue(int maxN){
	queue *q = (queue *)malloc(sizeof(queue));
	q->array = (int *)malloc(sizeof(int) * maxN);
	q->last = 0;
	q->first = 0;
	return q;
}

int element_num_queue(queue *q){
	return q->first - q->last;
}

void add_data_queue(int val, queue *q){
	q->array[q->first] = val;
	q->first++;
}

int take_data_queue(queue *q){
	if(element_num_queue(q) <= 0){
		printf("no data in the queue\n");
	}
	q->last++;
	return q->array[q->last - 1];
}

void flush_queue(queue *q){
	q->last = 0;
	q->first = 0;
}

int min(int a, int b){
	return a <= b ? a : b;
}

//n:頂点数, s:始点の番号, t:終点の番号, e:(u, v)->f uからvへの容量
int Edmonds_Karp(int N, int s, int t, int **e){
	int v, i, flow_sum = 0;
	int *flow = (int *)malloc(sizeof(int) * N);//各頂点に流せるフロー
	int *p = (int *)malloc(sizeof(int) * N);//増加道の履歴 前の頂点を入れる
	queue *q = make_queue(N * N);

	for(;;){
		//初期化
		for(i = 0; i < N; i++){
			flow[i] = 0;
			p[i] = -1;
		}
		flow[s] = inf;
		p[s] = -2;

		add_data_queue(s, q);
		while(element_num_queue(q) > 0){
			v = take_data_queue(q);
			if(v == t){
				break;
			}
			for(i = 0; i < N; i++){
				if(e[v][i] > 0 && p[i] == -1){
					p[i] = v;
					flow[i] = min(flow[v], e[v][i]);
					add_data_queue(i, q);
				}
			}
		}
		//増加道がなかった場合
		if(flow[t] == 0){
			break;
		}
		//残余ネットワークの構築
		for(v = t; v != s; v = p[v]){
			e[p[v]][v] -= flow[t];
			e[v][p[v]] += flow[t];
		}
		flow_sum += flow[t];
		flush_queue(q);
	}
	return flow_sum;
}

int main(){
	int N, i, j, ans;
	scanf("%d", &N);
	int **E = (int **)malloc(sizeof(int *) * (2 * N + 2));
	for(i = 0; i < 2 * N + 2; i++){
		E[i] = (int *)malloc(sizeof(int) * (2 * N + 2));
		for(j = 0; j < 2 * N + 2; j++){
			E[i][j] = 0;
		}
	}
	for(i = 1; i <= N; i++){
		E[0][i] = 1;
	}
	for(i = N + 1; i <= 2 * N; i++){
		E[i][2 * N + 1] = 1;
	}
	int *a = (int *)malloc(sizeof(int) * N);
	int *b = (int *)malloc(sizeof(int) * N);
	int *c = (int *)malloc(sizeof(int) * N);
	int *d = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d%d", &a[i], &b[i]);
	}
	for(i = 0; i < N; i++){
		scanf("%d%d", &c[i], &d[i]);
	}
	for(i = 0; i < N; i++){
		for(j = 0; j < N; j++){
			if(a[i] < c[j] && b[i] < d[j]){
				E[1 + i][N + 1 + j] = 1;
			}
		}
	}
/*	for(i = 0; i < 2 * N + 2; i++){
		for(j = 0; j < 2 * N + 2; j++){
			printf("E[%d][%d] = %d\n", i, j, E[i][j]);
		}
	}
*/	ans = Edmonds_Karp(2 * N + 2, 0, 2 * N + 1, E);
	printf("%d\n", ans);
	return 0;
}