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>
#define FOR(i,k,n)  for (int i=(k); i<(int)(n); ++i)
#define REP(i,n)    FOR(i,0,n)
#define GI(i) (scanf("%d",&i))

int N;
int a[200000],b[200000];

int main(void){
    GI(N);
    REP(i,N)GI(a[i]);
    REP(i,N)GI(b[i]);
    int r = 0;
    int s,t1,t2,t3,t4,t5,t6,t7,t8;
    for(int i=0;i<N;i++){
        s = a[i];
        for(int j=0;j+7<N;j+=8){
            t1 = s + b[j];
            t2 = s + b[j+1];
            t3 = s + b[j+2];
            t4 = s + b[j+3];
            t5 = s + b[j+4];
            t6 = s + b[j+5];
            t7 = s + b[j+6];
            t8 = s + b[j+7];
            t1 = t1 ^ t2;
            t3 = t3 ^ t4;
            t5 = t5 ^ t6;
            t7 = t7 ^ t8;
            t1 = t1 ^ t3;
            t5 = t5 ^ t7;
            r ^= t1 ^ t5;
        }
        for(int j=N/8*8;j<N;j++)
            r ^= (a[i] + b[j]);
    }
    printf("%d\n",r);
    return 0;
}