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;
#define FOR(i,a,b) for(int (i)=(a);i<(int)(b);i++)
#define rep(i,n) FOR(i,0,n)

struct Point {
    int x, y;
    bool operator<(const Point& r) const {
        return (x == r.x) ? (y < r.y) : (x < r.x);
    }
};

int main() {
    int n;
    scanf("%d", &n);
    vector<Point> a(n), b(n);
    rep(i, n) scanf("%d%d", &a[i].x, &a[i].y);
    rep(i, n) scanf("%d%d", &b[i].x, &b[i].y);
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    int ans = 0;
    static bool used[114] = {0};
    rep(i, n) {
        for (int j = n - 1; j >= 0; j--) {
            if (!used[j] && (a[j].x < b[i].x && a[j].y < b[i].y)) {
                used[j] = true;
                ans++;
                break;
            }
        }
    }

    printf("%d\n", ans);
    return 0;
}