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 <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#define ull unsigned long long
using namespace std;

const int nmax = 200100;
ull N;
ull a[nmax], b[nmax];

void read(){
    cin >> N;
    for(int i=0;i < N; ++i)cin >> a[i];
    for(int i=0;i < N; ++i)cin >> b[i];
}
ull bs(vector<ull>& bmod, ull a, ull l, ull r)
{
    if(l > (a + bmod[N-1]) || r < (a + bmod[0])){
        //if(a ==1)
        //    cout << l << ' ' << (a + bmod[N-1]) << ' ' << r << ' ' << r << ( a + bmod[0]) << endl;
        return 0;
    }
    ull L=0, R=N-1, mid;
    while((R-L)> 1)
    {
        mid = (L + R) >> 1;
        if((a + bmod[mid]) >= l)
            R = mid;
        else
            L = mid;
    }
    ull ans1;
    if(( a + bmod[L])< l)
        ans1 = R;
    else 
        ans1 = L;
    L = 0, R = N-1;
    while((R-L) > 1)
    {
        mid = (L + R) >> 1;
        if( ( a + bmod[mid]) > r)
            R = mid;
        else
            L = mid;
    }
    if((a + bmod[R]) > r)
        mid = L;
    else
        mid = R;
    //cout << a << ' ' << mid << ' ' << ans1 << endl;
    return mid - ans1 + 1;
}

void solve(){
    ull curmod = 2, ans = 0;
    for(int i = 0; i < 28; ++i)
    {
        vector <ull> bmod;
        for(int j=0; j < N; ++j)
            bmod.push_back(b[j] % curmod);
        sort(bmod.begin(), bmod.end());
        ull cnt = 0;
        for(int j=0;j<N; ++j)
        {
            cnt += bs(bmod, a[j] % curmod, (curmod >> 1), curmod - 1) + bs(bmod, a[j]  % curmod, (curmod >> 1) * 3, (curmod << 1) -1);
            //cout << cnt << ' ' << (1 << i) << endl;
        }
        ans += (cnt & 1) * (curmod >> 1);
        curmod <<=1;
    }
    cout << ans << endl;
}

int main(){
    read();
    solve();
}