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

int main(void)
{
  int n;
  long int x,y,maxsum,minsum,maxdif,mindif,ans;
  long int sum[200000];
  long int dif[200000];

  scanf("%d",&n);
  for(int i=0;i<n;i++)
  {
    scanf("%ld %ld",&x,&y);
    sum[i]=x+y;
    dif[i]=x-y;
    if(i==0)
    {
      maxsum=sum[i];
      minsum=sum[i];
      maxdif=dif[i];
      mindif=dif[i];
    }
    if(maxsum<sum[i]) maxsum=sum[i];
    if(minsum>sum[i]) minsum=sum[i];
    if(maxdif<dif[i]) maxdif=dif[i];
    if(mindif>dif[i]) mindif=dif[i];
  }

  if((maxsum-minsum)>(maxdif-mindif)) ans=maxsum-minsum;
  else ans=maxdif-mindif;

  printf("%ld\n",ans);

  return 0;
}