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<stdio.h>

int queue[100];
int front=0;
int rear=0;
int parent[100];
int f[202][202];

int search(int nn,int num,int **rc,int goal){
  int i,j,flag=1,flag2;
  if(rc[nn][goal]!=0){
    parent[goal]=nn;
    return 1;
  }
  for(i=0;i<num-1;i++){
    if(rc[nn][i]!=0){
	flag=1;
	for(j=0;j<rear;j++){
	  if(i==queue[j])  flag=0;
	}
	if(flag){
	  queue[rear]=i;
	  rear++;
	  parent[i]=nn;
	}
    }
  }
  
  flag2=0;
  front++;
  while(front<rear&&flag2==0){
    flag2= search(queue[front],num,rc,goal);
    front++;
  }
  return flag2;
}

int main(){
  int n,i,j,temp,a[100],b[100],c[100],d[100],flag[100]={},sum=0,max=0,m[202][202]={};
  int flow,*rc[202];
  scanf("%d",&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=1;i<=n;i++) {
    m[0][i]=1;
    m[i+n][2*n+1]=1;
  }
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      if(a[i]<c[j]&&b[i]<d[j]) m[i+1][n+j+1]=1;
    }
  }

  front=0;
  queue[0]=0;
  rear=1;

  for(i=0;i<2*n+2;i++) rc[i]=m[i];

  flow=search(0,2*n+2,rc,2*n+1);

  while(flow!=0){
    for(i=0;i<(2*n+2);i++)
	for(j=0;j<(2*n+2);j++)
	  f[i][j]=0;
    i=(2*n+2)-1;
    do{
	rc[parent[i]][i]-=flow;
	rc[i][parent[i]]+=flow;
	i=parent[i];
    }while(i!=0);
    sum+=flow;
    front=0;
    queue[0]=0;
    rear=1;
    flow=search(0,2*n+2,rc,2*n+1);
  }
  printf("%d\n",sum);
}