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<string.h>
#include<math.h>
#include<stdlib.h>
#include<limits.h>

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

typedef struct node{
    int x;
    int y;
    int f;
}node;

int cmpAscX(const void* a, const void* b){
    int x = ((node*)a)->x, y = ((node*)b)->x;
    if(x > y){
        return 1;
    }
    if(x < y){
        return -1;
    }
    return 0;
}

int cmpDscY(const void* a, const void* b){
    int x = ((node*)a)->y, y = ((node*)b)->y;
    if(x < y){
        return 1;
    }
    if(x > y){
        return -1;
    }
    return 0;
}

int main(){
    int n;
    scanf("%d", &n);
    node red[n];
    node blu[n];
    rep(i, 0, n){
        int x, y;
        scanf("%d %d", &x, &y);
        red[i].x = x;
        red[i].y = y;
        red[i].f = 1;
    }
    rep(i, 0, n){
        int x, y;
        scanf("%d %d", &x, &y);
        blu[i].x = x;
        blu[i].y = y;
        blu[i].f = 1;
    }
    int ans = 0;
    qsort(red, n, sizeof(node), cmpDscY);
    qsort(blu, n, sizeof(node), cmpAscX);
    rep(i, 0, n){
        rep(j, 0, n){
            if(red[j].x < blu[i].x && red[j].y < blu[i].y && red[j].f == 1){
                red[j].f = 0;
                ans++;
                break;
            }
        }
    }
    printf("%d", ans);

    return 0;
}