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: Wrong Answer)

import java.util.Scanner;
class Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long n=sc.nextLong();
		long[] al=new long[(int)n];
		long[] al1=new long[(int)n];
		long max,max1=0;	
		for(int i=0;i<n;i++)
		{
			al[i]=sc.nextLong();
			al1[i]=sc.nextLong();
			
		}
		for(int i=0;i<n-1;i++)
		{
			max=((Math.abs(Long.valueOf(al[(int)i]-al[(int)i+1])))+(Math.abs(Long.valueOf(al1[(int)i]-al1[(int)i+1]))));
			max1=Math.max(max, max1);
		}
		System.out.println(max1);
	}

}