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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

#define SENTINEL 1000000000000000000

void merge(long long A[], long long B[], long long left, long long mid, long long right){
  long long n1, n2, i, j, k;
  long long *L,*R, *Lb, *Rb;
  
  n1 = mid-left;
  n2 = right-mid;
  
  L = (long long*)malloc(sizeof(long long) *(n1 + 1));
  R = (long long*)malloc(sizeof(long long) *(n2 + 1));
  Lb = (long long*)malloc(sizeof(long long) *(n1 + 1));
  Rb = (long long*)malloc(sizeof(long long) *(n2 + 1));
  
  for(i = 0; i <= n1 - 1; i++){
    L[i] = A[left + i];
    Lb[i] = B[left + i];
  }
  for(j = 0; j <= n2 - 1; j++){
    R[j] = A[mid + j];
    Rb[j] = B[mid + j];
  }
  
  L[n1] = SENTINEL;
  R[n2] = SENTINEL;
  Lb[n1] = SENTINEL;
  Rb[n2] = SENTINEL;
  
  i = 0;
  j = 0;
  
  for(k = left; k <= right - 1; k++){
   if(L[i] <= R[j]){
      A[k] = L[i];
      B[k] = Lb[i];
      i++;
    }
    else{
      A[k] = R[j];
      B[k] = Rb[j];
      j++;
    }
  }
  free(L);
  free(R);
  free(Lb);
  free(Rb);
}

void mergeSort(long long A[], long long B[], long long left, long long right){
  long long mid;
  
  if(left + 1 < right){
    mid=(left + right)/2;
    mergeSort(A,B, left, mid);
    mergeSort(A,B, mid, right);
    merge(A,B, left, mid, right);
  }
}

int main(void){

    int N;

    scanf("%d", &N);

    long long a[N], b[N];

    for(int i = 0; i < N; i++){
        scanf("%lld%lld", &a[i], &b[i]);
    }

    long long c[N], d[N];

    for(int i = 0; i < N; i++){
        scanf("%lld%lld", &c[i], &d[i]);
    }

    mergeSort(a, b, 0, N);
    mergeSort(c, d, 0, N);

    int R = 0, B = 0;
    int l = 0;

    while(1){
        if(a[R] >= c[B]){
            B++;
        }else if(b[R] >= d[B]){
            R++;
        }else{
            l++;
            B++;
            R++;
        }
        if(B >= N || R >= N){
            break;
        }
    }

    mergeSort(b, a, 0, N);
    mergeSort(d, c, 0, N);

    int r = 0;

    R = 0, B = 0;

    while(1){
        if(b[R] >= d[B]){
            R++;
        }else if(a[R] >= c[B]){
            B++;
        }else{
            r++;
            R++;
            B++;
        }
        if(R >= N || B >= N){
            break;
        }
    }

    if(r < l){
        printf("%d\n", l);
    }else{
        printf("%d\n", r);
    }

    return 0;
}