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<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#define Rint register int
#define Temp template<typename T>
#define mem(a,b) memset(a,(b),sizeof(a))
typedef long long LL;
using namespace std;
Temp inline void read(T &x){
    x=0;T w=1,ch=getchar();
    while(!isdigit(ch)&&ch!='-')ch=getchar();
    if(ch=='-')w=-1,ch=getchar();
    while(isdigit(ch))x=(x<<3)+(x<<1)+(ch^'0'),ch=getchar();
    x=x*w;
}

const int maxn=1000+10;
int n;
int a[maxn],b[maxn],c[maxn],d[maxn];
int ans;
bool G[maxn<<1][maxn<<1];
bool vis[maxn];
int mat[maxn];
inline bool dfs(int x){
	for(Rint i=n+1;i<=(n<<1);i++){
		if(vis[i]==0&&G[x][i]){
			vis[i]=1;
			if(mat[i]==0||dfs(mat[i])){
				mat[i]=x;
				return true;
			}
		}
	}
	return false;
}

int main(){
	read(n);
	for(Rint i=1;i<=n;i++)read(a[i]),read(b[i]);
	for(Rint i=1;i<=n;i++)read(c[i]),read(d[i]);
	for(Rint i=1;i<=n;i++){
		for(Rint j=1;j<=n;j++){
			if(a[i]<c[j]&&b[i]<d[j])G[i][n+j]=1,G[n+j][i]=1;
		}
	}
	for(Rint i=1;i<=n;i++){
		mem(vis,0);
		if(dfs(i))ans++;
	}
	printf("%d\n",ans);
//	for(Rint i=n+1;i<=(n<<1);i++)cout<<i<<' '<<mat[i]<<endl;
    return 0;
}