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: Accepted)

#include <bits/stdc++.h>

#define int long long
#define double long double
#define ff first
#define ss second
#define endl '\n'
#define ii pair<int, int>
#define mp make_pair
#define mt make_tuple
#define DESYNC                      \
  ios_base::sync_with_stdio(false); \
  cin.tie(0);                       \
  cout.tie(0)
#define pb push_back
#define vi vector<int>
#define vii vector<ii>
#define all(x) x.begin(), x.end()
#define EPS 1e-9
#define INF 1e18
#define ROOT 1
#define M 1000000007
#define curtime chrono::steady_clock::now().time_since_epoch().count
#define rep(i, beg, n, s) for (int i = beg; i < n; i += s)
const double PI = acos(-1);

using namespace std;

inline int mod(int n, int m = M) {
  int ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}

int exp(int n, int k) {
  if (k == 0) return 1;
  if (k == 1) return n;
  int ax = exp(n, k / 2);
  ax = mod(ax * ax);
  if (k % 2) ax = mod(ax * n);
  return ax;
}

int gcd(int a, int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}

int calc(vector<ii>& v) {
  int ans = 0;
  for (ii p : v) {
    ans = max(ans, abs(p.ff - v[0].ff) + abs(p.ss - v[0].ss));
  }
  return ans;
}
//#define MULTIPLE_TEST_CASE
void solution() {
  int n;
  cin >> n;
  vector<ii> v(n);
  int max_x = -INF, min_x = INF, max_y = -INF, min_y = INF;
  for (ii& p : v) {
    int x, y;
    cin >> x >> y;
    max_x = max(max_x, x + y);
    min_x = min(min_x, x + y);
    max_y = max(max_y, x - y);
    min_y = min(min_y, x - y);
    p.ff = x + y;
    p.ss = x - y;
  }

  cout << max(max_x - min_x, max_y - min_y) << endl;
}

int32_t main() {
  DESYNC;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int t = 1;
#ifdef MULTIPLE_TEST_CASE
  cin >> t;
#endif
  while (t--) solution();
}