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)

/*

                         YATIN KWATRA                   _03_08_14


Let's Hoop..

------------------------------------------------

           o
 /|   o         o
 \|=--            o
    ##
                    \\
                 /   \O
                O_/   T
                T    /|
                |\  | |
 _______________|_|________

------------------------------------------------

o- - -  \o          __|
   o/   /|          vv`\
  /|     |              |
   |    / \_            |
  / \   |               |
 /  |                   |


------------------------------------------------

HOOPER_4_LYF..

*/

// AC BAKSHI RABBA
/*

                         YATIN KWATRA                   _03_08_14


Let's Hoop..

------------------------------------------------

           o
 /|   o         o
 \|=--            o
    ##
                    \\
                 /   \O
                O_/   T
                T    /|
                |\  | |
 _______________|_|________

------------------------------------------------

o- - -  \o          __|
   o/   /|          vv`\
  /|     |              |
   |    / \_            |
  / \   |               |
 /  |                   |


------------------------------------------------

HOOPER_4_LYF..

*/

// AC BAKSHI RABBA

// #pragma GCC optimize "trapv"

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define ll                      long long
#define int                     ll
#define FIO                     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ld                      long double
#define ull                     unsigned long long
#define endl                    "\n"
#define fo(i,a,b)               for(int i = a; i<=b ; i++)
#define rfo(i,a,b)              for(int i = a; i>=b ; i--)
#define vii                     vector<int>
#define vll                     vector<ll>
#define pq                      priority_queue
#define uom                     unordered_map
#define all(v)                  v.begin(),v.end()
#define mp                      make_pair
#define pb                      push_back
#define pob                     pop_back
#define ff                      first
#define ss                      second
#define pii                     pair<int,int>
#define pll                     pair<ll,ll>
#define mll                     map<ll,ll>
#define mii                     map<int,int>
#define vvii                    vector<vii>
#define vvll                    vector<vll>
#define mod                     1000000007
#define MIN                     -1e9
#define pi                      3.1415926535897932384626433832795

using namespace std;
using namespace __gnu_pbds;

template<class T> using oset =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;


void INPUT()
{
#ifndef ONLINE_JUDGE
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
#endif
}

/*
----------------------------------------------------------------

   -> Check For Overflows
   -> Out of Bounds
   -> Initialisations of global arrays and variables
   -> Read the Question and the sample test cases carefully
   -> Its not Always Complex, Try to simplify
   -> write it down and try to solve
   -> Have Faith in urself and in ur approach
   -> If it seems totally out of reach, switch to next :)
   -> And Come Back later, if possible ..

----------------------------------------------------------------*/

const int N = 2e5 + 5;

int n, x, y;

void solve()
{
  cin >> n;

  int ans = 0;

  int mn1 = 1e18, mx1 = -1e18, mn2 = 1e18, mx2 = -1e18, cur;
  pii pmn1, pmx1, pmn2, pmx2;

  while (n--)
  {
    cin >> x >> y;

    cur = x + y;

    if (cur < mn1)
    {
      pmn1 = {x, y};
      mn1 = cur;
    }

    if (cur > mx1)
    {
      pmx1 = {x, y};
      mx1 = cur;
    }

    /*    cur = llabs(x - y);

        if (cur < mn2)
        {
          pmn2 = {x, y};
          mn2 = cur;
        }

        if (cur > mx2)
        {
          pmx2 = {x, y};
          mx2 = cur;
        }*/
  }

  //int d1 = llabs(pmn1.ff - pmx1.ff) + llabs(pmn1.ss - pmx1.ss);
  //int d2 = llabs(pmn2.ff - pmx2.ff) + llabs(pmn2.ss - pmx2.ss);

// ans = max(d1, d2);
  ans = llabs(mn1 - mx1);
  cout << ans;
}

signed main() {

  FIO
  INPUT();

  int t;
  t = 1;
// cin >> t;

  fo(i, 0, t - 1)
  {
    solve();
  }
  return 0;
}