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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <ctype.h>
#include <inttypes.h>

//int compare_int(const void *a, const void *b){//qsort 整数比較<
//	return *(int*)a - *(int*)b;
//}
//int compare_int2(const void *a, const void *b){//qsort 整数比較>
//	return *(int*)b - *(int*)a;
//}
//

//long long int combination(int n, int r)
//{
//	if (r == 0 || r == n)
//		return (1);
//	else if (r == 1)
//		return (n);
//	return (combination(n - 1, r - 1) + combination(n - 1, r));
//}

int main(void){
	int n;
	int x[200010]={0},y[200010]={0};
	long long int ans[200010]={0};
	int i;
	long long int min=1000000000000000000,max=0;

	scanf("%d",&n);
	for(i=0;i<n;i++){
		scanf("%d %d",&x[i],&y[i]);
		ans[i]=x[i]+y[i];
	}
	for(i=0;i<n;i++){
		if(min>ans[i]){
			min=ans[i];
		}
		if(max<ans[i]){
			max=ans[i];
		}
	}
	printf("%lld",max-min);

	return 0;
}