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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define all(x) (x).begin(),(x).end()
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;}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin>>n;
    vector<P> ab, cd;
    rep(i,n){
        int a,b; cin>>a>>b;
        ab.push_back(make_pair(a,b));
    }
    rep(i,n){
        int a,b; cin>>a>>b;
        cd.push_back(make_pair(a,b));
    }
    sort(all(ab)); sort(all(cd));
    int ans = 0;

    // 条件に合うやつの中で一番yが近いやつをえらぶ
    multiset<int> m;
    const int INF = 1e9;
    int ind = 0; ans = 0;
    for(auto x : cd){
        while(ind < ab.size() && ab[ind].first < x.first) {
            m.insert(ab[ind].second);
            ind++;
        }
        auto it = m.lower_bound(x.second);
        if(it != m.begin()){
            ans++;
            m.erase(prev(it));
        }
    }

    cout<<ans<<endl;
}