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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#define ms_valtype int

#pragma GCC optimize("O3")
#pragma GCC target("avx")

//比較関数
int compare_ms(ms_valtype a, ms_valtype b){
	return a - b;
}

//昇順
void sort_sub(ms_valtype *origin, int left, int right, ms_valtype *tmp){
	if(right - left > 1){
		int i, j, k, half = (left + right) / 2;
		sort_sub(origin, left, half, tmp);
		sort_sub(origin, half, right, tmp);
		for(i = left; i < right; i++){
			tmp[i] = origin[i];
		}
		for(i = left, j = left, k = half; i < right; i++){
			if(k == right){
				origin[i] = tmp[j];
				j++;
			}
			else if(compare_ms(tmp[j], tmp[k]) <= 0 && j < half){
				origin[i] = tmp[j];
				j++;
			}
			else{
				origin[i] = tmp[k];
				k++;
			}
		}
	}
}

void sort(ms_valtype *origin, int N){
	ms_valtype *tmp = (ms_valtype *)malloc(sizeof(ms_valtype) * N);
	sort_sub(origin, 0, N, tmp);
	free(tmp);
}

int bound(int k, int N, int *b){
	int l = -1, h, r = N;
	while(r - l > 1){
		h = (l + r) / 2;
		if(b[h] < k){
			l = h;
		}
		else{
			r = h;
		}
	}
	return r;
}

int main(){
	int N, i, j, ans = 0, xor, dig, cov;
	scanf("%d", &N);
	int *a = (int *)malloc(sizeof(int) * N);
	int *b = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		scanf("%d", &a[i]);
	}
	for(i = 0; i < N; i++){
		scanf("%d", &b[i]);
	}
	for(i = 28; i >= 0; i--){
		dig = 1 << i, cov = (dig << 1) - 1;
		xor = N & 1;
//		printf("i = %d\n", i);
		for(j = 0; j < N; j++){
			a[j] &= cov;
			b[j] &= cov;
		}
		sort(b, N);
/*		for(j = 0; j < N; j++){
			printf("b[%d] = %d\n", j, b[j]);
		}
*/		for(j = 0; j < N; j++){
			xor ^= bound(1 * dig - a[j], N, b) & 1;
			xor ^= bound(2 * dig - a[j], N, b) & 1;
			xor ^= bound(3 * dig - a[j], N, b) & 1;
		}
//		printf("xor = %d\n", xor);
		ans |= (xor << i);
	}
	printf("%d\n", ans);
	return 0;
}