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)

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
const ll maxn=500005;
const ll mod=1e9+7;

int main(){
	fast;
	ll n,i,j,x,y,k;
	cin>>n;
	vector<pair<ll,ll>> vx(n),vy(n);
   for(i=0;i<n;i++){
	   cin>>x>>y;
	   vx[i]={x,y};
	   vy[i]={y,x};
   }	   
   sort(all(vx));
   sort(all(vy));
   ll ans=0;
   for(i=0;i<n;i++){
	   ans=max({ans,abs(vx[i].ff-vx[n-1].ff)+abs(vx[i].ss-vx[n-1].ss),abs(vx[i].ff-vx[0].ff)+abs(vx[i].ss-vx[0].ss),abs(vx[i].ff-vy[n-1].ss)+abs(vx[i].ss-vy[n-1].ff),abs(vx[i].ff-vy[0].ss)+abs(vx[i].ss-vy[0].ff)});
   }
   for(i=0;i<n;i++){
	   ans=max({ans,abs(vy[i].ff-vy[n-1].ff)+abs(vy[i].ss-vy[n-1].ss),abs(vy[i].ff-vy[0].ff)+abs(vy[i].ss-vy[0].ss),abs(vy[i].ff-vx[n-1].ss)+abs(vy[i].ss-vx[n-1].ff),abs(vy[i].ff-vx[0].ss)+abs(vy[i].ss-vx[0].ff)});
   }
   cout<<ans;
}