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;

#include<ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template <typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const int mod = (int)1e7+9;
#define REP0(i,n) for(int i=0;i<n;i++)
#define REP(i,a,n) for(int i=a;i<=n;i++)
#define F first
#define S second
#define endl "\n"
#define ll long long
#define pii pair<int,int>
#define vi vector<int>
#define maxpq priority_queue<int>
#define minpq priority_queue<int, vector<int>, greater<int> >
#define setbits(x) __builtin_popcountll(x)
#define inf (int)(1e18)
#define prec(x) fixed<<setprecision(x)<<x
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define all(x) x.begin(),x.end()

const int d4i[4]={-1, 0, 1, 0}, d4j[4]={0, 1, 0, -1};
const int d8i[8]={-1, -1, 0, 1, 1, 1, 0, -1}, d8j[8]={0, 1, 1, 1, 0, -1, -1, -1};

void solve(){
   int n;cin>>n; 
   vector<pii> points(n);
   REP0(i,n){
       cin>>points[i].first;
       cin>>points[i].second;
   }
   sort(all(points),[](pii A,pii B){return abs(A.F + A.S) < abs(A.S + B.S);}); 
   int ans = abs(abs(points[0].F + points[0].S) - abs(points[n-1].F + points[n-1].S));
   cout<<ans<<endl;
}


int main(){
    fastio;
    #ifndef ONLINE_JUDGE
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);
    #endif
    int tc = 1;
    // cin>>tc;
    for(int i=1;i<=tc;i++){
        solve();
    }
    return 0;
}