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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

console.log((args=>{
  const [[N],...tmp] = args.trim().split('\n').map(r=>r.split(' ').map(v=>v|0));
  const rp = [], bp = [];
  for ( let i = 0; i < N; i++ ) {
    rp.push(tmp[i]);
    bp.push(tmp[N+i]);
  }
  bp.sort(([a],[b])=>a-b);
  let count = 0;
  const test = [];
  for ( let i = 0; i < N; i++ ) {
    const [bx,by] = bp[i];
    let max = [-1,-1];
    for ( let j = 0; j < rp.length; j++ ) {
      const [rx,ry] = rp[j];
      if ( bx < rx ) continue;
      if ( by > ry && ry > max[1] ) max = [j,ry];
    }
    if ( max[1] >= 0 ) {
      rp.splice( max[0], 1 );
      count++;
    }
  }
  return `${count}`;
})(require('fs').readFileSync('/dev/stdin', 'utf8')));