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>
using namespace std;
#define F first
#define S second
#define R cin>>
#define ll long long
#define ln cout<<'\n'
#define in(a) insert(a)
#define pb(a) push_back(a)
#define pd(a) printf("%.10f\n",a)
#define mem(a) memset(a,0,sizeof(a))
#define all(c) (c).begin(),(c).end()
#define iter(c) __typeof((c).begin())
#define rrep(i,n) for(ll i=(ll)(n)-1;i>=0;i--)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define rep(i,n) REP(i,0,n)
#define tr(it,c) for(iter(c) it=(c).begin();it!=(c).end();it++)
ll check(ll n,ll m,ll x,ll y){return x>=0&&x<n&&y>=0&&y<m;}void pr(){ln;}
template<class A,class...B>void pr(const A &a,const B&...b){cout<<a<<(sizeof...(b)?" ":"");pr(b...);}
template<class A>void PR(A a,ll n){rep(i,n)cout<<(i?" ":"")<<a[i];ln;}
const ll MAX=1e9+7,MAXL=1LL<<61,dx[8]={-1,0,1,0,-1,-1,1,1},dy[8]={0,1,0,-1,-1,1,1,-1};
typedef pair<ll,ll> P;
typedef pair<ll,P> PP;

void Main() {
  ll n;
  R n;
  P a[n];
  rep(i,n) cin >> a[i].F >> a[i].S;
  ll ans=0;
  rep(k,4) {
    vector<PP> p(n);
    rep(i,n) {
      p[i].F=a[i].F+a[i].S;
      p[i].S=a[i];
    }
    sort(all(p));
    rep(i,1000) {
      if(i>=n) break;
      rep(j,1000) {
        if(n-1-j<0) break;
        P x=p[i].S,y=p[n-1-j].S;
        ans=max(ans,abs(x.F-y.F)+abs(x.S-y.S));
      }
    }
    rep(i,n) {
      swap(a[i].F,a[i].S);
      a[i].F=MAX-a[i].F;
    }
  }
  pr(ans);
}

int main(){ios::sync_with_stdio(0);cin.tie(0);Main();return 0;}