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>
#define sz(v)   ((int)(v).size())
#define  all(v)    ((v).begin()),((v).end())
#define  allr(v)    ((v).rbegin()),((v).rend())
#define   pb         push_back
#define   mp         make_pair
#define    clr(v,d)      memset( v, d ,sizeof(v))
#define     PI                       acos(-1)
typedef  long long     ll ;
typedef  unsigned long long ull;
const double eps= (1e-9);
using namespace std;
int dcmp(long double a,long double b){   return fabsl(a-b)<=eps ? 0: (a>b)? 1:-1  ;}
int getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1;}
int setBit1(int num, int idx) {return num | (1<<idx);}
ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}
ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}
void GO(){ ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);}
int countNumBit1(int mask) {int ret=0; while (mask) 	{mask &= (mask-1);++ret;	}return ret;}
const int N=200000+9,mod=1e9+7;
multiset<ll> S;
ll a[N],b[N];
int main(){
    GO();

    int T;
  //  cin>>T;
    T=1;
    while(T--){
        int n;
        cin>>n;
        S.clear();
        for(int i=0;i<n;i++){

            cin>>a[i]>>b[i];
            S.insert(a[i]+b[i]);
        }
        ll ans=0;
        for(int i=0;i<n;i++){
            S.erase(S.find(a[i]+b[i]));
            ans= max(ans,1ll*a[i]+b[i] - *(S.begin()));
            S.insert(a[i]+b[i]);
        }
        cout<<ans<<"\n";
    }

}