#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<ll, ll>
#define x first
#define y second

ii a[101], b[101];
ll n, ans = 0;
bool used[101];

bool ss(ii a, ii b){
    return (b.y > a.y);
}

int main(){
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i].x >> a[i].y;
    for (int i = 0; i < n; i++) cin >> b[i].x >> b[i].y;
    sort(a, a+n, greater<ii>());
    sort(b, b+n, ss);
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            if (a[i].x < b[j].x && a[i].y < b[j].y && !used[j]){
                ans++;
                used[j] = true;
                break;
            }
        }
    }
    cout << ans;
}
### 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<bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<ll, ll>
#define x first
#define y second

ii a[101], b[101];
ll n, ans = 0;
bool used[101];

bool ss(ii a, ii b){
    return (b.y > a.y);
}

int main(){
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i].x >> a[i].y;
    for (int i = 0; i < n; i++) cin >> b[i].x >> b[i].y;
    sort(a, a+n, greater<ii>());
    sort(b, b+n, ss);
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            if (a[i].x < b[j].x && a[i].y < b[j].y && !used[j]){
                ans++;
                used[j] = true;
                break;
            }
        }
    }
    cout << ans;
}
```