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;

using ll = long long;
using P = pair<int, int>;

const int INF = 1e9;
const int MOD = 1e9 + 7;

// 4近傍、8近傍
int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

bool used[110];

int main() {
  // 入力
  int N;
  cin >> N;
  vector<P> R;
  int a, b;
  for (int i = 0; i < N; i++) {
    cin >> a >> b;
    R.push_back(P(b, a));
  }
  vector<P> B;
  for (int i = 0; i < N; i++) {
    cin >> a >> b;
    B.push_back(P(a, b));
  }

  sort(R.begin(), R.end());
  sort(B.begin(), B.end());
  int ans = 0;
  for (int i = 0; i < N; i++) {
    int mini = 1000;
    int idx = -1;
    for (int j = 0; j < N; j++) {
      if (used[j]) continue;

      if (R[i].first < B[j].second && R[i].second < B[j].first) {
        if (mini > B[j].second) {
          mini = B[j].second;
          idx = j;
        }
      }
    }
    if (idx != -1) {
      ans++;
      used[idx] = true;
    }
  }

  // 解答
  cout << ans << endl;

  return 0;
}