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>

int main(void){
	int N;
	int red[100][3],blue[100][3];
	int i,j;
	
	scanf("%d",&N);
	for(i=0;i<N;i++){
		scanf("%d%d",&red[i][0],&red[i][1]);
	}
	for(i=0;i<N;i++){
		scanf("%d%d",&blue[i][0],&blue[i][1]);
	}
	for(i=0;i<N;i++){
		red[i][2]=red[i][0]+red[i][1];
		blue[i][2]=blue[i][0]+blue[i][1];
	}
//	printf("%d\n",red[0][2]);
	for(i=N-1;i>0;i--){
		for(j=0;j<i;j++){
			if(red[j][2]>red[j+1][2]){
				int x,y,z;
				x=red[j][0];
				y=red[j][1];
				z=red[j][2];
				red[j][0]=red[j+1][0];
				red[j][1]=red[j+1][1];
				red[j][2]=red[j+1][2];
				red[j+1][0]=x;
				red[j+1][1]=y;
				red[j+1][2]=z;
			}
		}
	}
	for(i=N-1;i>0;i--){
		for(j=0;j<i;j++){
			if(blue[j][2]>blue[j+1][2]){
				int x,y,z;
				x=blue[j][0];
				y=blue[j][1];
				z=blue[j][2];
				blue[j][0]=blue[j+1][0];
				blue[j][1]=blue[j+1][1];
				blue[j][2]=blue[j+1][2];
				blue[j+1][0]=x;
				blue[j+1][1]=y;
				blue[j+1][2]=z;
			}
		}
	}
	int memo[100];
	int k;
	int count=0;
	for(i=0;i<N;i++){
		for(j=0;j<N;j++){
			if(red[N-i][0]<blue[j][0] && red[N-i][1]<blue[j][1]){
				for(k=0;k<count;k++){
					if(j==memo[k]){
						break;
					}
				}
				memo[count]=j;
				count++;
				
				break;
			}
		}
	}
	if(count==0){
		printf("0");
	}else{
		printf("%d",count);
	}
	return 0;}