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: Accepted)

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=405;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
struct Edge{
    int to,nex;
}edge[N*N];
int n,m,head[N],tot,res[N],vist[N];
int a[N],b[N],c[N],d[N];
inline void Add(int u,int v)
{
    edge[++tot].to=v;
    edge[tot].nex=head[u];
    head[u]=tot;
}
bool dfs(int u)
{
    for(int i=head[u];i;i=edge[i].nex)
    {
        int v=edge[i].to;
        if(!vist[v])
        {
            vist[v]=1;
            if(!res[v]||dfs(res[v])){
                res[v]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int tmp;
    n=read();
    for(int i=1;i<=n;i++) a[i]=read(),b[i]=read();
    for(int i=1;i<=n;i++) c[i]=read(),d[i]=read();
    for(int i=1;i<=n;i++)
      for(int j=1;j<=n;j++)
        if(a[i]<c[j]&&b[i]<d[j])
          Add(i,j);
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        clr(vist,0);
        if(dfs(i))
          ans++;
    }
    printf("%d\n",ans);
    return 0;
}