// C - 2D Plane 2N Points
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(int)(n);++i)

int main(){
	int N; cin>>N;
	vector<vi> AB(N, vi(2)), CD(N, vi(2));
	rep(i, N) cin>>AB[i][0]>>AB[i][1];
	rep(i, N) cin>>CD[i][0]>>CD[i][1];

	sort(all(CD));
	sort(all(AB), [](vi a, vi b){ return a[1] > b[1]; });

	vector<vi> dim_b(N, vi());
	rep(b, N)
		rep(r, N)
			if(AB[r][0] < CD[b][0] && AB[r][1] < CD[b][1])
				dim_b[b].push_back(r);

	vi vis_r(N, false);
	rep(b, N)
		for(int r:dim_b[b])
			if(!vis_r[r]){ vis_r[r] = true; break; }

	cout<< accumulate(all(vis_r), 0) << endl;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

Examples

Input

3
2 0
3 1
1 3
4 2
0 4
5 5


Output

2


Input

3
0 0
1 1
5 2
2 3
3 4
4 5


Output

2


Input

2
2 2
3 3
0 0
1 1


Output

0


Input

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7


Output

5


Input

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9


Output

4

### Response

```cpp
// C - 2D Plane 2N Points
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(int)(n);++i)

int main(){
	int N; cin>>N;
	vector<vi> AB(N, vi(2)), CD(N, vi(2));
	rep(i, N) cin>>AB[i][0]>>AB[i][1];
	rep(i, N) cin>>CD[i][0]>>CD[i][1];

	sort(all(CD));
	sort(all(AB), [](vi a, vi b){ return a[1] > b[1]; });

	vector<vi> dim_b(N, vi());
	rep(b, N)
		rep(r, N)
			if(AB[r][0] < CD[b][0] && AB[r][1] < CD[b][1])
				dim_b[b].push_back(r);

	vi vis_r(N, false);
	rep(b, N)
		for(int r:dim_b[b])
			if(!vis_r[r]){ vis_r[r] = true; break; }

	cout<< accumulate(all(vis_r), 0) << endl;
}

```