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>
#include <iostream>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define speed                   \
  ios_base::sync_with_stdio(0); \
  cin.tie(0);                   \
  cout.tie(0);
#define ll long long
#define pb push_back
#define mem1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define endl "\n"
#define mod 1000000007
#define mod1 998244353
#define ff first
#define ss second
#define MAX 500005
#define N 200005
#define INF 1000000009
#define all(v) v.begin(), v.end()
#define sbit(a) __builtin_popcount(a)
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef pair<ll, ll> pll;
typedef pair<pll, ll> ppl;
typedef map<ll, ll> mpll;
typedef map<vector<ll>, ll> mpvl;
ll power(ll x, ll y, ll p)
{
  ll res = 1;

  x = x % p;

  if (x == 0)
    return 0;
  while (y > 0)
  {

    if (y & 1)
      res = (res * x) % p;

    y = y >> 1;
    x = (x * x) % p;
  }

  return res;
}

void solve()
{
  ll n;
  cin>>n;
  vector<pll> p(n);
  for(ll i=0;i<n;i++)
  {
    cin>>p[i].ff>>p[i].ss;
  }
  sort(all(p));
  ll ans=0;
  for(ll i=1;i<n;i++)
  {
      ll get=abs(p[0].ff-p[i].ff)+abs(p[i].ss-p[0].ss);
      ans=max(ans,get);
  }
  cout<<ans<<endl;
}
int main()
{ 
  speed;
  ll kk;
  kk = 1;
  //cin >> kk;
  while (kk--)
  {
    solve();
  }
  return 0;
}