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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.lang.*;
import java.io.*;

class Main{
	public static void main(String[] args) {
	  Scanner sc=new Scanner(System.in);
      int n=sc.nextInt();
      int[][] arr=new int[n][2];
      for(int i=0;i<n;i++){
          arr[i][0]=sc.nextInt();
          arr[i][1]=sc.nextInt();
      }
      int dist=0;
      for(int i=0;i<n;i++){
          for(int j=i+1;j<n;j++){
            int temp=Math.abs(arr[i][0]-arr[j][0])+Math.abs(arr[i][1]-arr[j][1]);
            dist=Math.max(dist,temp);
          }
      }
      System.out.println(dist);
	}
}