Problem:
Score : 500 points

Problem Statement

There are N points on the 2D plane, i-th of which is located on (x_i, y_i).
There can be multiple points that share the same coordinate.
What is the maximum possible Manhattan distance between two distinct points?

Here, the Manhattan distance between two points (x_i, y_i) and (x_j, y_j) is defined by |x_i-x_j| + |y_i-y_j|.

Constraints

2 \leq N \leq 2 \times 10^5

1 \leq x_i,y_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the answer.

Sample Input 1

3
1 1
2 4
3 2

Sample Output 1

4

The Manhattan distance between the first point and the second point is |1-2|+|1-4|=4, which is maximum possible.

Sample Input 2

2
1 1
1 1

Sample Output 2

0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <cstdio>
#include <algorithm>
#include <complex>

using namespace std;

using ll = long long int;


const int MAXN=2e5+7;
int N;

complex<int> arr[MAXN];

int comp1(const complex<int>& a,const complex<int>& b){
    return (a.real()+a.imag())<(b.real()+b.imag());
}

int main(){
    scanf("%d",&N);
    for(int i=0;i<N;i++) {
        int x,y;
        scanf("%d%d",&x,&y);
        arr[i]=complex<int>(x,y);
    }

    int ans=0;
    auto largest=max_element(arr,arr+N,comp1),least=min_element(arr,arr+N,comp1);
    auto temp=largest->imag()+largest->real()-least->real()-least->imag();
    ans=max(temp,ans);

    for(int i=0;i<N;i++) {
        int x=arr[i].real(),y=arr[i].imag();
        arr[i]=complex<int>(1000000001-x,y);
    }

    largest=max_element(arr,arr+N,comp1),least=min_element(arr,arr+N,comp1);
    temp=largest->imag()+largest->real()-least->real()-least->imag();
    ans=max(temp,ans);
    
    printf("%d\n",ans);


    return 0;
}