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: Time Limit Exceeded)

#include<stdio.h>
int x[300000],y[300000];
int abs(int k){
	if (k<0) return -k; else return k;
}
int dis(int i,int j){
	return abs(x[i]-x[j])+abs(y[i]-y[j]);
}
void qsx(int l,int r){
	int i,j,temp,mid;
	i=l;j=r;
	mid=(x[i]+x[j])/2;
	while (i<=j){
		while (x[i]<mid) i++;
		while (x[j]>mid) j--;
		if (i<=j){
			temp=x[i];
			x[i]=x[j];
			x[j]=temp;
			temp=y[i];
			y[i]=y[j];
			y[j]=temp;
			i++;j--;
		}
	}
	if (l<j) qsx(l,j);
	if (i<r) qsx(i,r);
}
void qsy(int l,int r){
	int i,j,temp,mid;
	i=l;j=r;
	mid=(y[i]+y[j])/2;
	while (i<=j){
		while (y[i]<mid) i++;
		while (y[j]>mid) j--;
		if (i<=j){
			temp=x[i];
			x[i]=x[j];
			x[j]=temp;
			temp=y[i];
			y[i]=y[j];
			y[j]=temp;
			i++;j--;
		}
	}
	if (l<j) qsy(l,j);
	if (i<r) qsy(i,r);
}
void qsxy(int l,int r){
	int i,j,temp,mid;
	i=l;j=r;
	mid=(x[i]+x[j]+y[i]+y[j])/2;
	while (i<=j){
		while (x[i]+y[i]<mid) i++;
		while (x[j]+y[j]>mid) j--;
		if (i<=j){
			temp=x[i];
			x[i]=x[j];
			x[j]=temp;
			temp=y[i];
			y[i]=y[j];
			y[j]=temp;
			i++;j--;
		}
	}
	if (l<j) qsxy(l,j);
	if (i<r) qsxy(i,r);
}
int main(){
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d %d",&x[i],&y[i]);
	}
	qsx(1,n);
	int ans=0;
	for (int i=1;i<=n;i++){
		if(dis(i,n)>ans)
		  ans=dis(i,n);
	}
	for (int i=1;i<=n;i++){
		if(dis(i,1)>ans)
		  ans=dis(i,1);
	}
	qsy(1,n);
	for (int i=1;i<=n;i++){
		if(dis(i,n)>ans)
		  ans=dis(i,n);
	}
	for (int i=1;i<=n;i++){
		if(dis(i,1)>ans)
		  ans=dis(i,1);
	}
	qsxy(1,n);
	for (int i=1;i<=n;i++){
		if(dis(i,n)>ans)
		  ans=dis(i,n);
	}
	for (int i=1;i<=n;i++){
		if(dis(i,1)>ans)
		  ans=dis(i,1);
	}
	printf("%d",ans);
	
	
	
}