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>

#define REP(i, n) for(int (i)=0;(i)<(n);++(i))
#define MOD 1000000007

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

bool pairCompare(const P& firstElof, const P& secondElof)
{
    return firstElof.second < secondElof.second;
}

int main()
{
    int N;
    cin >> N;
    vector<P> red, blue;
    REP(i, N)
    {   
        int x, y;
        cin >> x >> y;
        P p(x, y);
        red.push_back(p);
    }
    REP(i, N)
    {   
        int x, y;
        cin >> x >> y;
        P p(x, y);
        blue.push_back(p);
    }

    // x座標でソート
    sort(red.begin(), red.end());
    sort(blue.begin(), blue.end());
    int redAns = 0;
    bool isUsedX[N] = {false};
    REP(i, N)
    {
        int a, b;
        a = red[i].first;
        b = red[i].second;
        REP(j, N)
        {
            if (!isUsedX[j])
            {
                int c, d;
                c = blue[j].first;
                d = blue[j].second;
                if (a < c && b < d)
                {
                    redAns++;
                    isUsedX[j] = true;
                    break;
                }
            }
        }
    }

    // y座標でソート
    sort(red.begin(), red.end(), pairCompare);
    sort(blue.begin(), blue.end(), pairCompare);
    int blueAns = 0;
    bool isUsedY[N] = {false};
    REP(i, N)
    {
        int a, b;
        a = red[i].first;
        b = red[i].second;
        REP(j, N)
        {
            if (!isUsedY[j])
            {
                int c, d;
                c = blue[j].first;
                d = blue[j].second;
                if (a < c && b < d)
                {
                    blueAns++;
                    isUsedY[j] = true;
                    break;
                }
            }
        }
    }

    int ans = max(redAns, blueAns);

    cout << ans << endl;
}