#include<bits/stdc++.h>
using namespace std;
struct Node{
    int x,y;
}a[105],b[105];
int used[105],cnt[105];
int n;
int find(int x){
    for(int i=1;i<=n;i++){
        if((a[x].x<b[i].x)&&(a[x].y<b[i].y)&&!used[i]){
            used[i]=1;
            if(cnt[i]==0||find(cnt[i])){
                cnt[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i].x>>a[i].y;
    }
    for(int i=1;i<=n;i++){
        cin>>b[i].x>>b[i].y;
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(used,0,sizeof(used));
        if(find(i)){
            ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}

### Prompt

Please create a solution in Cpp 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
#include<bits/stdc++.h>
using namespace std;
struct Node{
    int x,y;
}a[105],b[105];
int used[105],cnt[105];
int n;
int find(int x){
    for(int i=1;i<=n;i++){
        if((a[x].x<b[i].x)&&(a[x].y<b[i].y)&&!used[i]){
            used[i]=1;
            if(cnt[i]==0||find(cnt[i])){
                cnt[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i].x>>a[i].y;
    }
    for(int i=1;i<=n;i++){
        cin>>b[i].x>>b[i].y;
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        memset(used,0,sizeof(used));
        if(find(i)){
            ans++;
        }
    }
    cout<<ans<<endl;
    return 0;
}

```