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>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define ll long long
#define PINT pair<int,int>
#define f1 first
#define s2 second
const int INF = 1e9 + 7;

using namespace std;

int main(){
    cin.tie(0), ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<PINT> red(n),blue(n);
    REP(i,n) cin >> red[i].f1 >> red[i].s2;
    REP(i,n) cin >> blue[i].f1 >> blue[i].s2;
    sort(red.begin(),red.end());
    sort(blue.begin(),blue.end());
    vector<bool> jud(n,false);
    int ans=0;
    REP(i,n){
        int max;
        bool flg=false;
        for(int j=0;red[j].f1<blue[i].f1&&j<n;j++){
            if(jud[j]) continue;
            if(red[j].s2 < blue[i].s2){
                if(!flg){
                    max=j;
                    flg=true;
                }
                if(red[max].s2 < red[j].s2) max = j;
            }
        }
        if(flg){
            jud[max] = true;
            ++ans;
        }
    }
    cout << ans << endl;
    return 0;
}