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: Wrong Answer)

#include <cstdio>
#include <string>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct point{
    int x,y,col,s;
}p[400];
int vis[400];
bool cmp(point a,point b){
    return a.s<b.s;
}
int main(){
    int N,M,res=0;
    scanf("%d",&N);
    for(int i=0;i<N;i++){
        cin>>p[i].x>>p[i].y;
        vis[i]=0;
        p[i].col=0;
        p[i].s=p[i].x*p[i].y;
    }
    for(int i=N;i<2*N;i++){
        cin>>p[i].x>>p[i].y;
        vis[i]=0;
        p[i].col=1;
        p[i].s=p[i].x*p[i].y;
    }
    sort(p,p+2*N,cmp);
    
    M=2*N-1;
    while(p[M].col==1)M--;
    for(int i=M;i>=0;i--){
        if(p[i].col==1)continue;
        for(int j=i+1;j<2*N;j++){
            if(vis[j]||!(p[j].col))continue;
            if(p[i].y<p[j].y&&p[i].x<p[j].x){
                res++;
                vis[i]=1;
                vis[j]=1;
                break;
            }
        }
    }
    printf("%d",res);
    return 0;
}