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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

var n = -1
var g = BooleanArray( 0 )
var e = Array( 0, { BooleanArray( 0 ) } )
var u = IntArray( 0 )

fun main( args: Array<String> ) {
    val sc = Scanner( System.`in` )
    n = sc.nextInt()
    val x1 = IntArray( n )
    val y1 = IntArray( n )
    val x2 = IntArray( n )
    val y2 = IntArray( n )
    for ( i in 0 until n ) {
        x1[i] = sc.nextInt()
        y1[i] = sc.nextInt()
    }
    for ( i in 0 until n ) {
        x2[i] = sc.nextInt()
        y2[i] = sc.nextInt()
    }

    e = Array( n, { i -> BooleanArray( n, { j -> x1[i] < x2[j] && y1[i] < y2[j] } ) } )
    u = IntArray( n, { -1 } )
    var r = 0
    for ( i in 0 until n ) {
        g = BooleanArray( n, { false } )
        if ( dfs( i ) ) {
            r ++
        }
    }
    println( r )
    System.out.flush()
}

fun dfs( v: Int ): Boolean {
    if ( v < 0 ) return true
    if ( g[v] ) return false
    g[v] = true
    for ( i in 0 until n ) {
        if ( e[v][i] && dfs( u[i] ) ) {
            u[i] = v
            return true
        }
    }
    return false
}