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: Time Limit Exceeded)

#include <limits.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define P 1000000007

int comp(const void *a, const void *b){return *(int*)a-*(int*)b;}
int compw(const void *a, const void *b){return (*(int*)a>*(int*)b)-(*(int*)a<*(int*)b);}

uint32_t nextpint(void){
	char c=getchar();
	while(c<'0'||'9'<c) c=getchar();
	uint_fast32_t x=0;
	while('0'<=c && c<='9'){ x=x*10+c-'0'; c=getchar(); }
	return x;
}

typedef struct {
	int to, cap, rev;
} edge;

edge e[202][202];
int l[202];
char u[202];

void add(int from, int to, int cap){
	int l1=l[from], l2=l[to];
	e[from][l1].to=to;
	e[from][l1].cap=cap;
	e[from][l1].rev=l2;
	e[to][l2].to=from;
	e[to][l2].cap=0;
	e[to][l2].rev=l1;
	l[from]++, l[to]++;
}

int dfs(int from, int to, int cap){
	if(from==to) return cap;
	u[from]=1;
	for(int i=0, j=l[from]; i<j; i++){
		int k=e[from][i].cap;
		if(cap>0 && u[e[from][i].to]==0){
			if(k>cap) k=cap;
			k=dfs(e[from][i].to, to, k);
			if(k>0){
				e[from][i].cap -= k;
				e[e[from][i].to][e[from][i].rev].cap += k;
				u[from]=0;
				return k;
			}
		}
	}
	u[from]=0;
	return 0;
}

int main(void){
	int n, i, j;
	int a[100][2], b[100][2];
	scanf("%d", &n);
	for(i=0; i<n; i++){ scanf("%d%d", &a[i][0], &a[i][1]); add(n*2, i, 1);}
	for(i=0; i<n; i++){ scanf("%d%d", &b[i][0], &b[i][1]); add(i+n, n*2+1, 1);}
	for(i=0; i<n; i++){
		for(j=0; j<n; j++){
			if(a[i][0]<b[j][0] && a[i][1]<b[j][1]) add(i, j+n, 1);
		}
	}
	int ans=0;
	while(1){
		int k=dfs(n*2, n*2+1, 1);
		if(k==0) break;
		ans+=k;
	}
	printf("%d\n", ans);
}