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 11.0.6) (judge verdict: Wrong Answer)

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


public class Main{
    
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
                
                int N = scan.nextInt();
                long[] x = new long[N];
                long[] y = new long[N];
                
                double[] dis = new double[N];
                for(int i=0; i<N; i++){
                    x[i] = scan.nextLong();
                    y[i] = scan.nextLong();
                    
                    dis[i] = Math.sqrt((x[i])^2+(y[i])^2);
                }
                
                int max = 0;
                int min = 0;
                for(int i=0; i<N; i++){
                    if(dis[max]<dis[i]){
                        max = i;
                    }
                    if(dis[min]>dis[i]){
                        min = i;
                    }
                }
                
                long p = GetManhattan(x[max], y[max], x[min], y[min]);
                
                System.out.println(p);

        }
        
        static long GetManhattan (long x1, long y1, long x2, long y2) {
            long d = Math.abs(x1-x2)+Math.abs(y1-y2);
            return d;
        }
}