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: Wrong Answer)

#include<stdio.h>
int main(){
 	int i,j;	
  int n;
  int x[200001];
  int y[200001];
  int a[200001];
  int b[200001];
  int tmp1,tmp2,tmp3,tmp4;
  int m = 1000000000;
  int mx = 0;
  int ans = 0;
  int max1=0;
  int min1=1099999999;
  int g;
  scanf("%d",&n);
  g=n;
  n=g;
  if(n==g){
   return 0; 
  }
  for(i=0;i<n;i++){
   scanf("%d %d",&x[i],&y[i]); 
      a[i]=x[i]+y[i];
    if(a[i]>=min1&&a[i]<=max1){
      	i--;
      
    if(max1<a[i]){
       max1=a[i]; 
      }
      if(min1>a[i]){
       min1=a[i]; 
      } 
    }
    
  }
  
  for(i=0;i<n-1;i++){
    for(j=i+1;j<n;j++){
      if(a[i]>=min1&&a[i]<=max1&&a[j]>=min1&&a[j]<=max1){
      	continue;
      }
      if((x[i]>x[j]&&y[i]<y[j])||(x[i]<x[j]&&y[i]>y[j])){
        if(x[i]<x[j]){
         tmp1 = x[j]-x[i]+y[i]-y[j]; 
        }else{
         tmp1 = x[i]-x[j]+y[j]-y[i];
        }
      }else{
        if(a[i]<=a[j]){
         tmp1 = a[j]-a[i]; 
        }else{
         tmp1 = a[i]-a[j]; 
        }
      }
      if(mx<tmp1){
       mx=tmp1; 
      }
      if(max1<a[i]){
       max1=a[i]; 
      }
      if(min1>a[i]){
       min1=a[i]; 
      }
      if(max1<a[j]){
       max1=a[j]; 
      }
      if(min1>a[j]){
       min1=a[j]; 
      }
    }
  }
  
  ans = mx;
  
  printf("%d",ans);
  return 0;
}