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>
int main(void){
    int n;
    scanf("%d",&n);
    long int t=0;
    long int s=0;
    long int x[200000]={};
    long int y[200000]={};
    int xmax=0;
    int xmin=0;
    int ymax=0;
    int ymin=0;
    long int ans=0;
    for(int i=0;i<n;i++){
        scanf("%ld %ld",&t,&s);
        x[i]=t+s;
        y[i]=t-s;
        //printf("%ld %ld\n",x[i],y[i]);
        
        if(!i){
            xmax=i;
            xmin=i;
        }else if(x[i]>x[xmax]){
            xmax=i;
        }else if(x[i]<x[xmin]){
            xmin=i;
        }
        if(!i){
            ymax=i;
            ymin=i;
        }else if(y[i]>y[ymax]){
            ymax=i;
        }else if(y[i]<y[ymin]){
            ymin=i;
        }
    }
    
    //printf("%ld %ld %ld %ld\n",x[xmax],x[xmin],y[ymax],y[ymin]);
    
    for(int i=0;i<n;i++){
        //printf(";%ld\n",(abs(x[i]-x[xmax])>=abs(y[i]-y[xmax])?abs(x[i]-x[xmax]):abs(y[i]-y[xmax])));
                                                                                                  //(abs(x[i]-x[xmax])>=abs(y[i]-y[xmax])?abs(x[i]-x[xmax]):abs(y[i]-y[xmax]))
        if(ans < (abs(x[i]-x[xmax])>=abs(y[i]-y[xmax])?abs(x[i]-x[xmax]):abs(y[i]-y[xmax]))  )ans=(abs(x[i]-x[xmax])>=abs(y[i]-y[xmax])?abs(x[i]-x[xmax]):abs(y[i]-y[xmax]));
        if(ans < (abs(x[i]-x[xmin])>=abs(y[i]-y[xmin])?abs(x[i]-x[xmin]):abs(y[i]-y[xmin]))  )ans=(abs(x[i]-x[xmin])>=abs(y[i]-y[xmin])?abs(x[i]-x[xmin]):abs(y[i]-y[xmin]));
        if(ans < (abs(x[i]-x[ymax])>=abs(y[i]-y[ymax])?abs(x[i]-x[ymax]):abs(y[i]-y[ymax]))  )ans=(abs(x[i]-x[ymax])>=abs(y[i]-y[ymax])?abs(x[i]-x[ymax]):abs(y[i]-y[ymax]));
        if(ans < (abs(x[i]-x[ymin])>=abs(y[i]-y[ymin])?abs(x[i]-x[ymin]):abs(y[i]-y[ymin]))  )ans=(abs(x[i]-x[ymin])>=abs(y[i]-y[ymin])?abs(x[i]-x[ymin]):abs(y[i]-y[ymin]));
    
        
    }
   printf("%ld\n",ans);
    
}