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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
#define int int64
#define rep(i, n) for(int i=0; i<n; i++)
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define SORT(x) sort(x.begin(), x.end())
#define GSORT(x) sort(x.begin(), x.end(), greater<int>())
#define mk make_pair
#define fi first
#define se second
#define pb push_back
#define ALL(x) x.begin(), x.end()
#define V(T) vector<T>
typedef pair<int, int> P;
typedef pair<P, P> PP;
typedef vector<int> vi;
typedef vector<vi> vvi;

bool check(P a, P b) {
    return a.fi+a.se > b.fi+b.se;
}

signed main() {
    int N;
    cin >> N;
    V(P) red(N), blue(N);
    rep(i, N) {
        int a, b;
        cin >> a >> b;
        red[i] = P(a, b);
    }
    rep(i, N) {
        int c, d;
        cin >> c >> d;
        blue[i] = P(c, d);
    }
    sort(ALL(red), check); sort(ALL(blue), check);
    //rep(i, N) cout << blue[i].fi << " " << blue[i].se << endl;
    vi done(N, 0);
  
    int res = 0;    
    rep(i, N) {
        rep(j, N) {
            if(blue[i].fi > red[j].fi && blue[i].se > red[j].se && !done[j]) {
               res++;
               done[j] = 1;
               break; 
            }
        }
    }

    cout << res << endl;
    return 0;
}