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: Wrong Answer)

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

#define int long long

void solve()
{
	int n;
	cin >> n;

	vector<int> x(n), y(n);
	for(int i = 0; i < n; i++)
		cin >> x[i] >> y[i];

	vector<int> add(n);
	for(int i = 0; i < n; i++)
		add[i] = abs(x[i] + y[i]);

	int res = 0;
	sort(add.begin(), add.end());
	res = abs(add.front() - add.back());

	vector<int> sub(n);
	for(int i = 0; i < n; i++)
		sub[i] = abs(x[i] - y[i]);

	sort(sub.begin(), sub.end());
	res = max(res, abs(sub.front() - sub.back()));

	cout << res << endl;

}

signed main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	solve();
	return 0;
}