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.*;
public class Main {
	public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int max = 0;
        List<List<Integer>> lists = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            List<Integer> list = new ArrayList<>();
            list.add(x);
            list.add(y);
            if (lists.size() == 0) {
                lists.add(list);
            }
            for (int j = 0; j < lists.size(); j++) {
                List<Integer> node = lists.get(j);
                int d = Math.abs(x - node.get(0)) + Math.abs(y - node.get(1));
                if (d > max) {
                    max = d;
                    lists.clear();
                    lists.add(list);
                    lists.add(node);
                }
            }

        }
        System.out.println(max);
	}
}