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: Runtime Error)

//
// Created by WUOG on 2020/9/13.
//


#include <cstdio>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;
double s[100009][6];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)scanf("%lf %lf",&s[i][0],&s[i][1]);
    double ans=0;
    for(int i=0; i<(1<<2); i++)//枚举每一种可能
    {
        double minn=inf,maxn=-inf;
        for(int j=0; j<n; j++)//计算每个数的值
        {
            double t=0;
            for(int k=0; k<2; k++)
            {
                if((1<<k)&i) t+=s[j][k];//采用位运算枚举
                else t-=s[j][k];
            }
            minn=min(minn,t);
            maxn=max(maxn,t);
        }
        ans=max(ans,(maxn-minn));//此即前i情况下对应的最大值
    }
    printf("%d\n",ans);
    return 0;
}