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;
        #define mod 1000000000000000000
        #define MOD 		1000000007
        #define ll long long
        #define ull unsigned ll
        #define mp 		make_pair
        #define pb 		push_back
        #define forn(i,n) 	for(ll i=0;i<n;i++)
        #define for1(i,n) 	for(ll i=1;i<=n;i++)
        #define forn1(i,n) 	for(int i=1;i<n;i++)
        #define rep(i,x,n)  for(ll i=x;i<=n;i++)
        #define repr(i,x,j) for(ll i=x;i>=j;i--)
        #define vi vector<ll>
        #define mii map<ll,ll>
        #define vii vector<pair<ll,ll>>
        #define vvi vector<vector<ll>>
        #define pii pair<ll,ll>
        #define F first
        #define S second
        #define fill(a,x) memset(a,x,sizeof a);
        #define all(x) x.begin(),x.end()
        #define test ll t;cin>>t;while(t--)
        #define yes cout<<"YES"<<endl
        #define no cout<<"NO"<<endl
        #define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
        ll flag[1000];
        #define R return 0
        int main()
        {
            IOS;
            vi x,y;
            ll n,a,b;
            cin>>n;
            forn(i,n)
            {
                cin>>a>>b;
                x.pb(a),y.pb(b);

            }
            vi v;
            ll i1=max_element(all(x))-x.begin();
            ll i2=min_element(all(x))-x.begin();
            ll i3=max_element(all(y))-y.begin();
            ll i4=min_element(all(y))-y.begin();
            //cout<<i1<<" "<<i2<<" "<<i3<<" "<<i4;
            ll d1=abs(x[i1]-x[i2])+abs(y[i1]-y[i2]);
            ll d2=abs(x[i3]-y[i4])+abs(x[i3]-y[i4]);
            v.pb(d1);
            v.pb(d2);
            ll d3=abs(x[i1]-x[i3])+abs(y[i1]-y[i3]);
            v.pb(d3);
            ll d4=abs(x[i1]-x[i4])+abs(y[i1]-y[i4]);
            v.pb(d4);
            ll d5=abs(x[i2]-x[i3])+abs(y[i2]-y[i3]);
            v.pb(d5);
            ll d6=abs(x[i2]-x[i4])+abs(y[i2]-y[i4]);
            v.pb(d6);
            sort(v.rbegin(),v.rend());
            cout<<v[0]<<endl;
            
        }
        
        /// Happy coding ):