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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define cinf(n,x) for(int i=0;i<(n);i++)cin>>x[i];
#define ft first
#define sc second
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(v) (v).begin(),(v).end()
#define mod 1000000007
using namespace std;
typedef long long ll;
template<class T> using V=vector<T>;
using Graph = vector<vector<int>>;
using P=pair<int,int>;
typedef unsigned long long ull;
typedef long double ldouble;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

const ll INF=1e18;

typedef pair<double,int> PP;

int main(){
    int n;
    cin>>n;
    V<P> r(n),b(n);
    rep(i,n){
        int a,t;
        cin>>a>>t;
        r[i]=P(a,t);
    }
    rep(i,n){
        int c,d;
        cin>>c>>d;
        b[i]=P(c,d);
    }
    sort(all(r),greater<P>());
    sort(all(b),greater<P>());
    int ans=0;
    set<int> st;
    for(int i=0;i<n;i++){
        V<PP> cnt;
        for(int j=0;j<n;j++){
            if(r[i].ft<b[j].ft&&r[i].sc<b[j].sc&&!st.count(j)){
                int x=r[i].ft-b[j].ft;
                int y=r[i].sc-b[j].sc;
                cnt.push_back(PP((double)sqrt(x*x+y*y),j));
            }
        }
        if(cnt.size()>0){
            sort(all(cnt));
            ans++;
            st.insert(cnt[0].sc);
        }
    }
    cout<<ans<<endl;
}