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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    val n = readLine()!!.toInt()
    val pos = Array(n) { readLine()!!.split(" ").map { it.toInt() } }

    var max = Math.abs(pos[1][0]-pos[0][0]) + Math.abs(pos[1][1]-pos[0][1])
    var bl = pos[0]
    var tr = pos[1]

    for (i in 2 until n) {
        val p = pos[i]

        if (p[0] >= Math.min(tr[0], bl[0]) && p[0] <= Math.max(tr[0], bl[0])
            && p[1] >= Math.min(tr[1], bl[1]) && p[1] <= Math.max(tr[1], bl[1])) continue

        val dbl = Math.abs(p[0]-bl[0]) + Math.abs(p[1]-bl[1])
        var dtr = Math.abs(p[0]-tr[0]) + Math.abs(p[1]-tr[1])

        if (dbl <= max && dtr <= max) continue

        if (dtr > dbl) {
            bl = p
            max = dtr
        } else {
            tr = p
            max = dbl
        }
    }

    println(max)
}