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>
const int INF = 1e9;
const int MOD = 1e9+7;
const long long LINF = 1e18;
#define dump(x)  cout << 'x' << ' = ' << (x) << ` `;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) for(int i=0;i<(n);++i)
#define REPR(i,n) for(int i=n;i>=0;i--)
#define FOREACH(x,a) for(auto& (x) : (a) )
typedef long long ll;
using namespace std;
typedef pair<ll, ll> P;

int main(int argc, char const *argv[]) {
    int n;
    cin >> n;
    vector<P> red, blue;
    vector<ll> used;
    REP(i,n) {
        int a,b;
        cin >> a >> b;
        red.emplace_back(a,b);
    }
    REP(i,n) {
        int c, d;
        cin >> c >> d;
        blue.emplace_back(c,d);
    }

    sort(red.begin(), red.end());
    sort(blue.begin(), blue.end());

    int ans = 0;
    // x座標が小さい順に見ていく
    vector<ll> ymax;
    REP(i,n) {
        REP(j,n) {
            if (red[j].first >= blue[i].first) break;
            if (blue[i].second >= red[j].second) {
                ymax.push_back(red[j].second);
            }
        }
        sort(ymax.begin(), ymax.end(), greater<>());
        sort(used.begin(), used.end());
        for (auto&& e:ymax) {
            if (!binary_search(used.begin(), used.end(), e)) {
                ++ans;
                used.push_back(e);
                break;
            }
        }
    }

    cout << ans << endl;

    return 0;
}