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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
#include<math.h>
#include<limits.h>
#define MAX 202
int max(int a,int b){
    if(a>b) return a;
    else return b;
}

int main(){
    int N;
    int a[MAX],b[MAX],c[MAX],d[MAX];
    bool red[MAX][MAX];
    memset(red,false,sizeof(red));
    bool blue[MAX][MAX];
    memset(blue,false,sizeof(blue));
    scanf("%d",&N);
    for(int i=1;i<=N;i++){
        scanf("%d %d",a+i,b+i);
        red[a[i]][b[i]]=true;
    }
    for(int i=1;i<=N;i++){
        scanf("%d %d",c+i,d+i);
        blue[c[i]][b[i]]=true;
    }
    int red_num[MAX][MAX],blue_num[MAX][MAX];
    for(int i=0;i<=2*N;i++)
        red_num[i][0] = red_num[0][i] = blue_num[i][0] = blue_num[0][i] = 0;        
    for(int i=1;i<=2*N;i++){
        for(int j=1;j<=2*N;j++){
            red_num[i][j] += max(red_num[i-1][j],red_num[i][j-1])+((red[i][j])?1:0);
            blue_num[i][j] += max(blue_num[i-1][j],blue_num[i][j-1])+((blue[i][j])?1:0); 
        }
    }
    int DP[MAX][MAX];
    memset(DP,0,sizeof(DP));
    for(int i=1;i<=2*N;i++){
        for(int j=1;j<=2*N;j++){
            DP[i][j] = max(DP[i-1][j],DP[i][j-1]) + (((blue_num[i][j]>=red_num[i][j]) && blue[i][j])?1:0);
        }
    }
    printf("%d",DP[2*N][2*N]);
}