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 ll long long
#define ld long double

using namespace std;

bool sortsec(const tuple<ll,ll,ll> &a, const tuple<ll,ll,ll> &b){

    return (get<1>(a) < get<1>(b));
}

bool sortthird(const tuple<ll,ll,ll> &a, const tuple<ll,ll,ll> &b){

    return (get<2>(a) < get<2>(b));
}

int main(){

    ll n;
    cin >> n;
    vector<tuple <ll, ll, ll>> xy;

    ll x, y;
    for(int i=0; i<n; i++){
        cin >> x >> y;
        xy.push_back({x+y, x, y});
    }

    sort(xy.begin(), xy.end());
    ll zero = abs(get<1>(xy[0]) - get<1>(xy[n-1])) + abs(get<2>(xy[0]) - get<2>(xy[n-1]));

    sort(xy.begin(), xy.end(), sortsec);
    ll one = abs(get<1>(xy[0]) - get<1>(xy[n-1])) + abs(get<2>(xy[0]) - get<2>(xy[n-1]));

    sort(xy.begin(), xy.end(), sortthird);
    ll two = abs(get<1>(xy[0]) - get<1>(xy[n-1])) + abs(get<2>(xy[0]) - get<2>(xy[n-1]));

    ll ans = max(zero, one);
    ans = max(ans, two);

    cout << ans << endl;
    
    return 0;
}