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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define SWAP(type, x, y) do { type tmp = x; x = y; y = tmp; } while (0)

#define MOD 1000000007 

typedef struct{
    int x;
    int y;
    int flg;
}data;

//x軸の昇順
int xasc(const void* a, const void* b){
    data p = *(data*)a;
    data q = *(data*)b;
    if(p.x!=q.x) return p.x-q.x;
    else return p.y-q.y;
}

//y軸の降順
int ydesc(const void* a, const void* b){
    data p = *(data*)a;
    data q = *(data*)b;
    if(p.y!=q.y) return q.y-p.y;
    else return q.x-p.x;
}

int main(void){
    int N;
    scanf("%d",&N);

    //A[0]~A[N-1]に格納する
    data* red;
    NEW(red,N);
    for(int i=0;i<N;i++){
        scanf("%d%d",&red[i].x,&red[i].y);
        red[i].flg=0;
    }
    qsort(red,N,sizeof(data),ydesc);

    data* blue;
    NEW(blue,N);
    for(int i=0;i<N;i++){
        scanf("%d%d",&blue[i].x,&blue[i].y);
        blue[i].flg=0;
    }
    qsort(blue,N,sizeof(data),xasc);

    int ans=0;
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if(red[j].flg) continue;
            if(red[j].x<blue[i].x && red[j].y<blue[i].y){
                ans++;
                red[j].flg=1;
                break;
            }
        }
    }
    printf("%d\n",ans);

    return 0;
}