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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

n = int(input())
xl, yl = [], []
for _ in range(n):
    x, y = map(int, input().split())
    xl.append(x)
    yl.append(y)
xl_ = xl.copy()
yl_ = yl.copy()
xl_.sort()
yl_.sort()

if n%2:
    xm, ym = xl_[n//2], yl_[n//2]
else:
    xm, ym = (xl_[n//2-1]+xl_[n//2])/2, (yl_[n//2-1]+yl_[n//2])/2

xm, ym = xm-1e-6, ym-1e-6
ans1, ans2, ans3, ans4 = 0, 0, 0, 0
for i in range(n):
    if xl[i] > xm:
        if yl[i] > ym:
            ans1 = max(ans1, abs(xl[i]-xm)+abs(yl[i]-ym))
        else:
            ans4 = max(ans4, abs(xl[i]-xm)+abs(yl[i]-ym))
    else:
        if yl[i] > ym:
            ans2 = max(ans2, abs(xl[i]-xm)+abs(yl[i]-ym))
        else:
            ans3 = max(ans3, abs(xl[i]-xm)+abs(yl[i]-ym))
ans = int(max(abs(ans1+ans3), abs(ans2+ans4)))

xm, ym = xm+1e-6+1e-6, ym+1e-6+1e-6
ans1, ans2, ans3, ans4 = 0, 0, 0, 0
for i in range(n):
    if xl[i] > xm:
        if yl[i] > ym:
            ans1 = max(ans1, abs(xl[i]-xm)+abs(yl[i]-ym))
        else:
            ans4 = max(ans4, abs(xl[i]-xm)+abs(yl[i]-ym))
    else:
        if yl[i] > ym:
            ans2 = max(ans2, abs(xl[i]-xm)+abs(yl[i]-ym))
        else:
            ans3 = max(ans3, abs(xl[i]-xm)+abs(yl[i]-ym)) 
ans_ = int(max(abs(ans1+ans3), abs(ans2+ans4)))
print(max(ans, ans_))