cf 57A

A. Square Earth?

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Meg the Rabbit decided to do something nice, specifically — to determine the shortest distance between two points on the surface of our planet. But Meg... what can you say, she wants everything simple. So, she already regards our planet as a two-dimensional circle. No, wait, it's even worse — as a square of side n. Thus, the task has been reduced to finding the shortest path between two dots on a square (the path should go through the square sides). To simplify the task let us consider the vertices of the square to lie at points whose coordinates are: (0, 0), (n, 0), (0, n) and (n, n).

Input

The single line contains 5 space-separated integers: n, x1, y1, x2, y2 (1 ≤ n ≤ 1000, 0 ≤ x1, y1, x2, y2 ≤ n) which correspondingly represent a side of the square, the coordinates of the first point and the coordinates of the second point. It is guaranteed that the points lie on the sides of the square.

Output

You must print on a single line the shortest distance between the points.

Sample test(s)

input

2 0 0 1 0

output

1

input

2 0 1 2 1

output

4

input

100 0 0 100 100

output

200

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
int n,a,b,c,d;
int main()
{
    int sum=0;
    scanf("%d",&n);
    scanf("%d%d%d%d",&a,&b,&c,&d);
    if((a<=b&&c<=d)||(a>=b&&c>=d)) printf("%d
",abs(a-c)+abs(b-d));
    else printf("%d
",min(4*n-a-b-c-d,a+b+c+d));
    return 0;
}