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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var i,j:longint;
    v:array[0..500]of boolean;
    goout,a,b,c,d:array[0..500]of longint;
    n:longint;
    ans:longint;
procedure bfs;
var min,i,x:longint;
begin
  repeat
    min:=n*4;
    for i:=1 to n do
    if (goout[n+i]<min)and(goout[n+i]>0) then
    begin
      min:=goout[n+i];
      x:=i;
    end;
    if min=n*4 then break;
    goout[n+x]:=n*4;
    for i:=1 to n do
    if not v[i] then
    begin
      if (c[x]>a[i])and(d[x]>b[i]) then
      begin
        v[i]:=true;
        for j:=1 to n do
        if (c[j]>a[i])and(d[j]>b[i])and(goout[n+j]<n*4) then dec(goout[n+j]);
        inc(ans);
        break;
      end;
    end;
  until min=n*4;
end;
begin
  read(n);
  for i:=1 to n do
  read(a[i],b[i]);
  for i:=1 to n do
  read(c[i],d[i]);
  for i:=1 to n do
    for j:=1 to n do
    if (a[i]<c[j])and(b[i]<d[j]) then
    begin
      inc(goout[n+j]);
    end;
  bfs;
  writeln(ans);
end.