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>
int ta[1010],to[1000010],nt[1000010],co[1000010],F[1010];
int MIN(int a,int b){return a<b?a:b;}
int MF(int s,int t,int e,int min){
  if(s==t||min==0)return min;
  int i,r;
  F[s]=1;
  for(i=ta[s];i+1;i=nt[i]){
    if(F[to[i]])continue;
    r=MF(to[i],t,e,MIN(min,co[i]));
    co[i]-=r;
    co[(i+e)%(2*e)]+=r;
    if(r)return r;
  }
  return 0;
}
//呼び出す方
//sからtへの最大流を求める　ノード数v辺の数e　aからbへキャパcのフローが流れる
int maxf(int s,int t,int v,int e,int *a,int *b,int *c){
  int i,f,r;
  for(i=0;i<v;i++)ta[i]=-1;
  for(i=0;i<e;i++){
    nt[i  ]=ta[to[i+e]=a[i]];
    nt[i+e]=ta[to[i  ]=b[i]];
    co[ta[a[i]]=i  ]=c[i];
    co[ta[b[i]]=i+e]=0;
  }
  for(f=0;r=MF(s,t,e,1000000000);f+=r){
    for(i=0;i<v;i++)F[i]=0;
  }
  return f;
}
int main(){
  int n,a[100010],b[100010],c[100010],r=0,i,j,x[1010],y[1010];
  scanf("%d",&n);
  for(i=0;i<2*n;i++)scanf("%d %d",&x[i],&y[i]);
  for(i=0;i<n;i++){
    a[r]=2*n;
    b[r]=i;
    c[r++]=1;
    a[r]=i+n;
    b[r]=2*n+1;
    c[r++]=1;
  }
  for(i=0;i<n;i++){
    for(j=n;j<2*n;j++){
      if(x[i]<x[j]&&y[i]<y[j]){
	a[r]=i;
	b[r]=j;
	c[r++]=1;
      }
    }
  }
  printf("%d\n",maxf(2*n,2*n+1,2*n+2,r,a,b,c));
  return 0;
}