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 ll long long
#define f first
#define s second
#define pb push_back
int main(){
    ll n;
    cin>>n;
    ll x,y;
    vector<pair<ll,ll>> vp;
    for(int i=0;i<n;i++){
        cin>>x>>y;
        vp.push_back({x,y});
    }
    ll ans=0;
    sort(vp.begin(),vp.end());
    // for(int i=0;i<n;i++){
    //     for(int j=i+1;j<n;j++){
    //         ans = max(ans, (abs(vp[i].f-vp[j].f) + abs(vp[i].s-vp[j].s)));
    //     }
    // }
    for(int i=0;i+1<n;i++){
        ans = max(ans, (abs(vp[0].f-vp[i+1].f) + abs(vp[0].s-vp[i+1].s)));
    }
    cout<<ans;
}