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<iostream>
#include <algorithm>
#include<string>
using namespace std;

int main()
{
	//int h, w,   q, D, ax[90001], ay[90001], l[100000], r[100000];
	long long int  i,j,k,l, a[1000], b[1000], c[1000], d[1000], e = 0, n, m, ten[1000] = { 0 }, ans = 0;
	string st[1000], stm[1000];

	cin >> n;
	for (i = 0; i < n; i++)
	{
		cin >> a[i];
		cin >> b[i];
		

	}for (i = 0; i < n; i++)
	{
		
		cin >> c[i];
		cin >> d[i];

	}
	for (i = 0; i < n; i++) {
		for (j = n - 1; j > i; j--) {
			if (a[j - 1] > a[j]) {  /* 前の要素の方が大きかったら */
				m = a[j];        /* 交換する */
				a[j] = a[j - 1];
				a[j - 1] = m;
				m = b[j];        /* 交換する */
				b[j] = b[j - 1];
				b[j - 1] = m;

			}
		}
	}
	//for (i = 0; i < n - 1; i++) {
	//	for (j = n - 1; j > i; j--) {
	//		if (b[j - 1] > b[j]) {  /* 前の要素の方が大きかったら */
	//			m = b[j];        /* 交換する */
	//			b[j] = b[j - 1];
	//			b[j - 1] = m;
	//		}
	//	}
	//}
	for (i = 0; i < n; i++) {
		for (j = n - 1; j > i; j--) {
			if (c[j - 1] > c[j]) {  /* 前の要素の方が大きかったら */
				m = c[j];        /* 交換する */
				c[j] = c[j - 1];
				c[j - 1] = m;
				m = d[j];        /* 交換する */
				d[j] = d[j - 1];
				d[j - 1] = m;
			}
		}
	}
	//for (i = 0; i < n - 1; i++) {
	//	for (j = n - 1; j > i; j--) {
	//		if (d[j - 1] > d[j]) {  /* 前の要素の方が大きかったら */
	//			m = d[j];        /* 交換する */
	//			d[j] = d[j - 1];
	//			d[j - 1] = m;
	//		}
	//	}
	//}
	for ( i = 0; i < n; i++)
	{
		k = 10000;
		for ( j = 0; j < n; j++)
		{
			if (a[i]<c[j])
			{
				if (b[i]<d[j])
				{
					if (k>d[j])
					{
						/*cout << a[i] << endl;
						cout << b[i] << endl;
						cout << c[j] << endl;
						cout << d[j] << endl;
						cout << "aaaaaaaaa" << endl;
						*/
						k = d[j];
						l = j;
					}
				}
			}
			/*else
			{
				break;
			}*/
		}
		if (k!=10000)
		{
			c[l] = -100000;
			d[l] = -100000;
			ans++;
		}
	}
		cout << ans << endl;

	
	// 出力
	return 0;
}