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.ArrayList;
import java.util.List;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
	    Scanner sc = new Scanner(System.in);
	    int n = sc.nextInt();

	    List<long[]> list = new ArrayList();

	    long maxx = 0;
	    long minx = 0;
	    long maxy = 0;
	    long miny = 0;
	    long returnval = 0;
	    for (int i = 0; i < n; i++) {
		    long x = sc.nextInt();
		    long y = sc.nextInt();
	    	if (i == 0) {
	    		maxx = x;
	    		minx = x;
	    		maxy = y;
	    		miny = y;
	    	}
	    	if (x < maxx && x > minx && maxy > y && miny < y) {
	    		continue;
	    	}
	    	maxx = Math.max(x, maxx);
	    	minx = Math.min(x, minx);
	    	maxy = Math.max(y, maxy);
	    	miny = Math.min(x, miny);
		    long[] wk = {x, y};
		    if (!list.contains(wk)) {
		    	if (list.size() >= 1) {
		    		for (int j = 0; j < list.size(); j++) {
		    			long[] wk2 = list.get(j);
		    			returnval = Math.max(Math.abs(wk[0] - wk2[0]) + Math.abs(wk[1] - wk2[1]), returnval);
		    		}
		    	}
		    	list.add(wk);
		    }
	    }
	    System.out.println(returnval);
	}

}