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 n:longint;
    i,k:longint;
    a,b,c,d:array[0..201000]of longint;
    m0,m1:longint;
    ans:int64;
procedure qs(l,r:longint);
var i,j,m,t:longint;
begin
  i:=l;
  j:=r;
  m:=c[(l+r)>>1];
  repeat
    while c[i]<m do inc(i);
    while c[j]>m do dec(j);
    if i<=j then
    begin
      t:=c[i];c[i]:=c[j];c[j]:=t;
      inc(i);
      dec(j);
    end;
  until i>j;
  if l<j then qs(l,j);
  if i<r then qs(i,r);
end;
function find(x:longint):longint;
var l,r,m:longint;
begin
  l:=1;
  r:=n;
  while l<=r do
  begin
    m:=(l+r)>>1;
    if c[m]<x then l:=l+1 else r:=m-1;
  end;
  exit(l);
end;

begin
  read(n);
  for i:=1 to n do
  read(a[i]);
  for i:=1 to n do
  read(b[i]);
  c[0]:=-1;
  c[n+1]:=maxlongint;
  for k:=1 to 29 do
  begin
    for i:=1 to n do
    begin
      c[i]:=a[i] and ((1 << k)-1);
      d[i]:=b[i] and ((1 << k)-1);
    end;
    qs(1,n);
    m0:=0;
    m1:=0;
    for i:=1 to n do
    begin
      m0:=m0+find((1 << (k-1))-d[i])-1;
      m0:=m0+find((1 << k)+(1 << (k-1))-d[i])-find((1 << k)-d[i]);
      m1:=m1+find((1 << k)-d[i])-find((1 << (k-1))-d[i]);
      m1:=m1+n+1-find((1 << k)+(1 << (k-1))-d[i]);
    end;
    //writeln('0=',m0,' 1=',m1);
    if m1 and 1 =1 then ans:=ans+1 << (k-1);
  end;
  writeln(ans);
end.