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)

#[allow(unused_imports)]
use itertools::Itertools;
#[allow(unused_imports)]
use num::*;
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::collections::*;

fn solve() {
    input! {
        n: usize,
        mut xyv: [(i64, i64); n]
    };

    let mut left = std::i64::MAX;
    let mut left_pos = (0, 0);
    let mut right = -1;
    let mut right_pos = (0, 0);
    let mut top = -1;
    let mut top_pos = (0, 0);
    let mut bottom = std::i64::MAX;
    let mut bottom_pos = (0, 0);
    for &(x, y) in &xyv {
        if x < left {
            left = x;
            left_pos = (x, y);
        }
        if x > right {
            right = x;
            right_pos = (x, y);
        }
        if y < bottom {
            bottom = y;
            bottom_pos = (x, y);
        }
        if y > top {
            top = y;
            top_pos = (x, y);
        }
    }
    let mut res = 0;
    for &(x, y) in &xyv {
        let dist1 = (left_pos.0 - x).abs() + (left_pos.1 - y).abs();
        let dist2 = (right_pos.0 - x).abs() + (right_pos.1 - y).abs();
        let dist3 = (bottom_pos.0 - x).abs() + (bottom_pos.1 - y).abs();
        let dist4 = (top_pos.0 - x).abs() + (top_pos.1 - y).abs();
        res = res.max(dist1).max(dist2).max(dist3).max(dist4);
    }
    println!("{}", res);
}

fn main() {
    std::thread::Builder::new()
        .name("big stack size".into())
        .stack_size(256 * 1024 * 1024)
        .spawn(|| {
            solve();
        })
        .unwrap()
        .join()
        .unwrap();
}