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: Time Limit Exceeded)

#include<bits/stdc++.h>
#include<vector>
#include<string>
#include<stack>
#include<iostream>
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define MOD 1000000007
#define vec(a) vector<int>a
using namespace std;


int main()
{
#ifndef ONLINE_JUDGE
	freopen("output.txt", "w", stdout);
	freopen("input.txt", "r", stdin);
#endif
	ll n,i,j,k,l,m,x,y;
	cin>>n;
	set< pair<ll,ll> > s;
	for(i=0;i<n;i++)
	{
		cin>>x>>y;
		s.insert(make_pair(x,y));
	}
	set< pair<ll,ll> >::iterator it,it1,it2;
	l=0;
	for(it=s.begin();it!=s.end();it++)
	{
		it2=it;
		it2++;
		for(it1=it2;it1!=s.end();it1++)
		{
			k=abs((*it1).first-(*it).first)+abs((*it1).second-(*it).second);
			if(k>l)
				l=k;
		}
	}

	cout<<l<<"\n";
	return 0;
}