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 <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <cmath>
#include <queue>
#include <set>
#include <stack>

using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define repr(i, n) for(ll i = n - 1; i >= 0; i--)
#define repone(i, n) for(ll i = 1; i <= (ll)(n); i++)
#define each(i, mp) for(auto i:mp)
#define FOR(i, m, n) for(ll i = m;i < n;i++)
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ll mod = 1000000007;
const ll inf = ll(1e9);
const ll half_inf = ll(1e5);
const ll ll_inf = ll(1e9) * ll(1e9);

int n;
vector<P> ab, cd;

void solve() {
    sort(ab.begin(), ab.end());
    sort(cd.begin(), cd.end());
    ll ans = 0;
    bool used[n];
    fill(used, used + n, false);
    repr(i, n) {
        P _xy = cd[i];
        repr(j, n) {
            P xy = ab[j];
            if (!used[j] && xy.first < _xy.first && xy.second < _xy.second) {
                used[j] = true;
                ans++;
                break;
            }
        }
    }
    cout << ans << endl;
}

int main() {
    cin >> n;
    ab.resize(n);
    cd.resize(n);
    int x, y;
    rep(i, n) {
        cin >> x >> y;
        ab[i] = make_pair(x, y);
    }
    rep(i, n) {
        cin >> x >> y;
        cd[i] = make_pair(x, y);
    }
    solve();
    return 0;
}