#include<iostream>
#include<string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdint>
#include <map>
#define MOD 1000000007
using namespace std;
typedef long long ll;

int main(){

int n;
cin >> n;

int r[201];
int b[201];

for(int i=0; i < 201; i++){
r[i] = -1;
b[i] = -1;
}

for(int i=0; i < n; i++){
int x;
cin >>x;
cin >> r[x];
}

for(int i=0; i < n; i++){
int x;
cin >>x;
cin >> b[x];
}



int ans = 0;
for(int i=0; i < n*2; i++){
if(b[i] != -1){
 int m = 200;   // a[200] = -1である 
 for(int j = 0; j < i; j++){
 if(r[m] < r[j] && r[j] < b[i]) m = j;
  }
 if(m != 200) { ans++; r[m] = -1;}
 }
}

cout << ans << endl;

return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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
#include<iostream>
#include<string>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstdint>
#include <map>
#define MOD 1000000007
using namespace std;
typedef long long ll;

int main(){

int n;
cin >> n;

int r[201];
int b[201];

for(int i=0; i < 201; i++){
r[i] = -1;
b[i] = -1;
}

for(int i=0; i < n; i++){
int x;
cin >>x;
cin >> r[x];
}

for(int i=0; i < n; i++){
int x;
cin >>x;
cin >> b[x];
}



int ans = 0;
for(int i=0; i < n*2; i++){
if(b[i] != -1){
 int m = 200;   // a[200] = -1である 
 for(int j = 0; j < i; j++){
 if(r[m] < r[j] && r[j] < b[i]) m = j;
  }
 if(m != 200) { ans++; r[m] = -1;}
 }
}

cout << ans << endl;

return 0;
}
```