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 rep(i, n) for(int i=0; i<n; i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
#define repr(i, n) for(int i=n-1; i>=0; i--)
#define repr1(i, n) for(int i=n; i>=1; i--)
#define all(v) v.begin(),v.end()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = 1e9+7;
const double EPS = 1e-10;
const double PI = acos(-1);
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; }

int main() {
    int n; cin >> n;
    vector<ll> x(n), y(n);
    rep(i, n) cin >> x[i] >> y[i];

    vector<ll> z(n), w(n);
    rep(i, n) {
        z[i] = x[i] + y[i];
        w[i] = x[i] - y[i];
    }

    ll zMax = -1, zMin = LLINF;
    ll wMax = -1, wMin = LLINF;
    rep(i, n) {
        chmax(zMax, z[i]); chmin(zMin, z[i]);
        chmax(wMax, w[i]); chmin(wMin, w[i]);
    }
    cout << max(zMax - zMin, wMax - wMin) << endl;
}