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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    public Main() {
    }

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
int N=scan.nextInt();
int  A[]=new int [N] ;
int  B[]=new int [N];

for(int i=0;i<N;i++){
    A[i]=scan.nextInt();
    B[i]=scan.nextInt();
}
int M=0;
for(int i=0;i<N;i++){
    for(int j=0;j<i;j++){
        if(M<Math.abs(A[i]-A[j])+Math.abs(B[i]-B[j])){M=Math.abs(A[i]-A[j])+Math.abs(B[i]-B[j]);}

    }

}
System.out.println(M);

    }
}