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 Python (3.8.2) (judge verdict: Wrong Answer)

N=int(input())
migisita=[0,10^9]
migiue=[0,0]
hidarisita=[10^9,10^9]
hidariue=[10^9,0]
for i in range(0,N):
    a,b=list(map(int,input().split()))
    if migiue[0]<a and migiue[1]<b:
        migiue=[a,b]
    if migisita[0]<a and migisita[1]>b:
        migisita=[a,b]
    if hidariue[0]>a and hidariue[1]<b:
        hidariue=[a,b]
    if hidarisita[0]>a and hidarisita[1]>b:
        hidarisita=[a,b]

one=abs(migiue[0]-hidarisita[0])+abs(migiue[1]-hidarisita[1])
two=abs(hidariue[0]-migisita[0])+abs(hidariue[1]-hidarisita[1])
three=abs(migiue[0]-migisita[0])+abs(migiue[1]-migisita[1])
four=abs(hidariue[0]-hidarisita[0])+abs(hidariue[1]-hidarisita[1])
print(max(one,two,three,four))