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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  
  vector<int> u(n), v(n);
  int up = -2e9, down = 2e9, left = 2e9, right = -2e9;
  for (int i = 0; i < n; i++) {
    int x, y;
    cin >> x >> y;
    u[i] = x + y;
    v[i] = x - y;
    up = max(up, v[i]);
    down = min(down, v[i]);
    left = min(left, u[i]);
    right = max(right, u[i]);
  }
  
  int ans = 0;
  for (int i = 0; i < n; i++) {
    ans = max({ans, abs(u[i] - left), abs(u[i] - right), abs(v[i] - up), abs(v[i] - down)});
  }
  
  cout << ans << '\n';
}