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 <iostream>
#include <sstream>
#include <stack>
#include <iomanip>
#include <numeric>
#include <string>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cstdlib>
#include <cassert>
#include <map>
#include <set>
#include <vector>
#include <utility>
#include <queue>
#include <cmath>

#define BEGIN(x) x.begin()
#define END(x) x.end()
#define ALL(x) BEGIN(x), END(x)
#define PAIR make_pair
#define VEC(type) vector<type >
#define endl '\n'
#define FIR first
#define SEC second
#define PB push_back
#define SZ(x) ((int)x.size())

#define FOR(i, A, B) for(auto i = (A);i != (B); i++)
#define FORD(i, A, B) for(auto i = (A);i != (B); i--)

#define READRANGE(begin, end) FOR(it, begin, end) cin >> *it
#define READVEC(V) READRANGE(BEGIN(V), END(V))

using namespace std;

typedef long long lint;
typedef pair<int, int> Pii;
typedef pair<int, lint> Pil;
typedef pair<lint, lint> Pll;
typedef pair<lint, int> Pli;

template<class U, class V>
istream& operator >>(istream& in, pair<U, V>& pr) {
    return in >> pr.first >> pr.second;
}

template<class U, class V>
ostream& operator <<(ostream& out, pair<U, V>& p) {
    return out << p.first << " " << p.second;
}

template<class T>
ostream& operator <<(ostream& out, vector<T> vec) {
    for (T& e : vec)
        out << e << " ";
    return out << endl;
}

// ---- BEGIN LIBRARY CODE ----
// ---- END LIBRARY CODE ----
void io_init()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
};

void solveSingleCase(int caseNo)
{
    int N;
    cin >> N;
    VEC(Pii) A(N), B(N);
    READVEC(A);
    sort(ALL(A));
    READVEC(B);
    sort(ALL(B));

    VEC(bool) used(N, false);
    int ret = 0;
    for (Pii b : B) {
        Pii take(-1, -1);
        FOR(i, 0, N)
            if (!used[i] && A[i].FIR < b.FIR && A[i].SEC < b.SEC) {
                take = max(take, Pii(A[i].SEC, A[i].FIR));
            }

        if (take.FIR < 0)
            continue;

        FOR(i, 0, N)
            if (!used[i] && Pii(A[i].SEC, A[i].FIR) == take) {
                used[i] = true;
                ret++;
                break;
            }
    }

    cout << ret << endl;
};

int main(int argc, char **argv)
{
    io_init();

    // Actual program code starts here.
    int T = 1;
    // cin >> T;
    FOR(tcase, 1, T + 1) {
        solveSingleCase(tcase);
    }
    return 0;
};