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: Time Limit Exceeded)



import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine().trim());
        Pair[] pairs=new Pair[n];
        for(int i=0;i<n;i++){
            String[] s=br.readLine().split(" ");
            int x=Integer.parseInt(s[0].trim());
            int y=Integer.parseInt(s[1].trim());
            pairs[i]=new Pair(x,y);
        }
        int res=Integer.MIN_VALUE;
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                int r=Math.abs(pairs[i].x-pairs[j].x)+Math.abs(pairs[i].y-pairs[j].y);
                res=Math.max(r,res);
            }
        }
        System.out.println(res);







    }


}
class Pair{
    int x,y;
    Pair(int x,int y){
        this.x=x;
        this.y=y;
    }
}