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;
typedef long long ll;
typedef unsigned long long LL;
typedef long double ld;
typedef vector<ll>VI;
typedef pair<ll,ll>P;
#define VV(T) vector<vector<T>>
#define sz(x) int(x.size())
#define PI 3.1415926535897932384626433832795
#define rep(i, n) for (ll i = 0; i < (ll)n; i++)
#define reprev(i,n) for (ll i=(ll)n-1;i>=0;i--)
#define ALL(a)  (a).begin(),(a).end()
#define rALL(a) (a).rbegin(),(a).rend()
#define c_max(a, b) (((ll)a)>((ll)b)?(a):(b))
#define c_min(a,b) (((ll)a)<((ll)b)?(a):(b))
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
#define vmax(v) *max_element(ALL(v))
#define vmin(v) *min_element(ALL(v))
#define $(x) {cout<<#x<<" = " <<(x)<<endl;}
#define fi first
#define se second
#define MAX 100100//5
#define MAX2 200100
#define MAX6 1001001//6
#define MAX7 10010010//7
#define SENTINEL 2000000000//9
#define INF 1<<30
#define INFTY 1LL<<62
#define MAX_INT INT_MAX
#define Endl '\n'
#define CLR(mat) memset(mat, 0, sizeof(mat))
inline ll GCD(ll a,ll b){return b?GCD(b,a%b):a;}//最大公約数
inline ll lcm(ll a,ll b){return a*b/GCD(a,b);}//最小公倍数
template<class T> bool contain(const std::string& s, const T& v) {
   return s.find(v) != std::string::npos;
}
const int MOD = 1000000007;

inline ll mod_add(ll a,ll b){return (a+b)%MOD;}
inline ll mod_sub(ll a,ll b){return (a+MOD-b)%MOD;}
inline ll mod_mul(ll a,ll b){return ((a%MOD)*(b%MOD))%MOD;}
inline ll mod_pow(ll x,ll n,ll mod=(ll)MOD){ll res=1;while(n>0){
if(n&1)res=res*x%mod;
x=x*x%mod;n>>=1;}return res;}
inline ll mod_inv(ll x){return mod_pow(x,MOD-2);}
inline ll mod_div(ll a, ll b){return mod_mul(a,mod_inv(b));}


//const int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
// vector型から重複を削除 list.erase(unique(ALL(list)),list.end());
// g++ -o a a.cpp -Wall -lm -std=c++17
//push_back -> emplace_back

int main(){
   ll n;cin>>n;
   vector<P> p(n);
   rep(i,n)cin>>p[i].fi>>p[i].se;
   vector<P> pp(n);
   rep(i,n){
      pp[i].fi=p[i].fi+p[i].se;
      pp[i].se=p[i].fi-p[i].se;
   }
   ll xmax=0,xmin=INFTY,ymax=0,ymin=INFTY;
   rep(i,n){
      chmax(xmax,pp[i].fi);
      chmin(xmin,pp[i].fi);
      chmax(ymax,pp[i].se);
      chmin(ymin,pp[i].se);
   }
   cout<<c_max(xmax-xmin,ymax-ymin)<<endl;
}