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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main(){
int N; 
cin >> N; 
int a[314], b[314], c[314], d[314]; 
for(int i = 0; i < N; i++){ 
cin >> a[i] >> b[i]; 
}
for(int i = 0; i < N; i++){ 
cin >> c[i] >> d[i]; 
}
int p[314], q[314]; 
for(int i = 0; i < N; i++){
p[i] = 201 * a[i] + b[i]; 
}
sort(p, p + N);
int x[314], y[314]; 
for(int i = 0; i < N; i++){
x[i] = p[i] / 201; 
y[i] = p[i] % 201; 
}
for(int i = 0; i < N; i++){
q[i] = 201 * d[i] + c[i]; 
}
sort(q, q + N);
int z[314], w[314]; 
for(int i = 0; i < N; i++){
w[i] = q[i] / 201; 
z[i] = q[i] % 201; 
}

int count = 0; 
for(int i = N-1; i >= 0; i--){
int L = upper_bound(w, w+N, y[i]) - w; 
for(int j = L; j < N; j++){
if(z[j] > x[i]) {
  z[j] = -1;
  count ++;
  break;  
  }
 }
}
cout << count << endl; 
}