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.Scanner;

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

			int[] x = new int[n];
			int[] y = new int[n];

			int minValue = Integer.MAX_VALUE;
			int minIndex = -1;
			int maxValue = Integer.MIN_VALUE;
			int maxIndex = -1;

			for (int i = 0; i < n; i++) {
				int a = x[i] = sc.nextInt();
				int b = y[i] = sc.nextInt();

				int u = a + b;
				if (u < minValue) {
					minValue = u;
					minIndex = i;
				}
				if (maxValue < u) {
					maxValue = u;
					maxIndex = i;
				}
			}
			
			System.out.println(Math.abs(x[minIndex]-x[maxIndex]) + Math.abs(y[minIndex]-y[maxIndex]));

		} catch (Exception e) {
			e.printStackTrace();
		}

	}

}