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>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PA;
typedef priority_queue<int, vector<int>, greater<int> > PQG;
#define REP(i, a, n) for(ll i = (ll) a;i < (ll) n;i++)
#define fi first
#define se second
#define pb push_back
#define INF 1000000009
#define MOD 1000000007
#define MAXN 103
#define MAXC 100000000000014
#define MAXV 1000000009

int n, dp[MAXN][MAXN], dp2[MAXN][MAXN];
PA ab[MAXN], cd[MAXN];

int main()
{
    cin >> n;
    REP(i, 0, n){
        cin >> ab[i].fi >> ab[i].se;
    }
    REP(i, 0, n){
        cin >> cd[i].fi >> cd[i].se;
    }
    sort(ab, ab + n);
    sort(cd, cd + n);
    REP(i, 0, n){
        REP(j, 0, n){
            if(cd[i].fi > ab[j].fi && cd[i].se > ab[j].se){
                dp[i + 1][j + 1] = max(dp[i][j] + 1, max(dp[i][j + 1], dp[i + 1][j]));
            }
            else{
                dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
            }
            //printf("dp[%lld][%lld] = %d\n", i + 1, j + 1, dp[i + 1][j + 1]);
        }
    }
    REP(i, 0, n){
        swap(cd[i].fi, cd[i].se);
        swap(ab[i].fi, ab[i].se);
    }
    sort(ab, ab + n);
    sort(cd, cd + n);
    REP(i, 0, n){
        REP(j, 0, n){
            if(cd[i].fi > ab[j].fi && cd[i].se > ab[j].se){
                dp2[i + 1][j + 1] = max(dp2[i][j] + 1, max(dp2[i][j + 1], dp2[i + 1][j]));
            }
            else{
                dp2[i + 1][j + 1] = max(dp2[i][j + 1], dp2[i + 1][j]);
            }
            //printf("dp[%lld][%lld] = %d\n", i + 1, j + 1, dp[i + 1][j + 1]);
        }
    }

    cout << max(dp[n][n], dp2[n][n]) << endl;



    return 0;
}