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 Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
    input! {
        n: usize,
        v: [(i64,i64); n]
    }

    let mut v_x = v.clone();
    v_x.sort_by(|t1,t2| t1.cmp(t2));
    let v_y = v.clone();
    v_y.iter().map(|(x,y)| (*y,*x)).collect::<Vec<(i64,i64)>>().sort_by(|t1,t2| t1.cmp(t2));
    let v_y = v_y.iter().map(|(x,y)| (*y,*x)).collect::<Vec<(i64,i64)>>();

    let mut hs0 = v_x[0];
    let mut hs1 = v_y[0];
    let mut hu0 = v_x[0];
    let mut hu1 = v_x[0];
    let mut ms0 = v_x[0];
    let mut ms1 = v_x[0];
    let mut mu0 = v_x[n-1];
    let mut mu1 = v_y[n-1];

    for i in 1..n {
        if v_x[i].0 != hs0.0 {
            hu0 = v_x[i-1];
        }
        if v_x[i].0 == mu0.0 {
            ms0 = v_x[i];
            break;
        }
    }

    for i in 1..n {
        if v_y[i].1 != hs1.1 {
            ms1 = v_y[i-1];
        }
        if v_y[i].1 == mu1.1 {
            hu1 = v_y[i];
            break;
        }
    }

    let vec = vec![hs0,hs1,hu0,hu1,ms0,ms1,mu0,mu1];
    let ans ={
        let mut ans = 0;
        for i in 0..8 {
            for j in 0..8 {
                let (x,y) = vec[i];
                let (z,w) = vec[j];
                ans = ans.max((x-z).abs()+(y-w).abs());
            }
        }
        ans
    };

    println!("{}",ans)
}