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)

#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; i++)
#define repr(i, n) for(int i=n; i>=0; i--)
#define reps(i, s, e) for(int i=s; i<e; i++)
#define inf 1e18
#define vsort(v) sort(v.begin(), v.end())
#define vsortr(v) sort(v.begin(), v.end(), greater<lli>())
#define sz(x) x.size()
#define ceil(a, b) (a+b-1)/b
#define ok cout << "ok" << endl;
using namespace std;
using ll = long long;
template<typename T> bool chmax(T &a, T &b){ if(a<b) { a=b; return 1; }}
template<typename T> bool chmin(T &a, T &b){ if(b<a) { a=b; return 1; }}
template<typename T> T gcd(T a, T b){ if(b==0) return a; return gcd(b, a%b); }
template<typename T> T lcm(T a, T b){ return a*b/gcd(a, b); }
template<typename T> T vdebug(vector<T> v){ for(auto vv : v){ cout << vv << " "; } cout << endl; }
template<typename T> T adebug(T arr[], int n){ rep(i, n){ cout << arr[i] << " "; } cout << endl; }
void ans(bool b){ if(b) cout << "Yes" << endl; else cout << "No" << endl; }
void ans2(bool b){ if(b) cout << "YES" << endl; else cout << "NO" << endl; }
int keta(ll num){ int k=0; while(num>0){ num/=10; k++; } return k; }
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};


int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, a, b, c, d;
    vector<pair<int, int>> red, blue;
    cin >> n;
    rep(i, n){
        cin >> a >> b;
        red.push_back(make_pair(a, b));
    }
    rep(i, n){
        cin >> c >> d;
        blue.push_back(make_pair(c, d));
    }

    sort(red.begin(), red.end());
    sort(blue.begin(), blue.end());

    int res = 0;
    int used[1000];
    rep(i, n){
        used[i] = 0;
    }
    for(auto const& r : red){
        rep(i, n){
            if(used[i]) continue;
            if(r.first >= blue[i].first) continue;
            if(r.second < blue[i].second){
                res++;
                used[i] = 1;
                break;
            }
        }
    }

    cout << res << endl;

    return 0;
}