Square Earth?

F - Square Earth?

Time Limit:2000MS     Memory Limit:262144KB    

Description

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 Input

Input

2 0 0 1 0

Output

1

Input

2 0 1 2 1

Output

4

Input

100 0 0 100 100

Output

200

此题可以说是道水题。理解题意就好。分三种情况。


代码：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int p,q,x,t;
    int n,x1,y1,x2,y2;
    cin>>n>>x1>>y1>>x2>>y2;
    p=abs(x1-x2);
    q=abs(y1-y2);
    if(p==n)                      //情况1
    {
        x=y1+y2;
        t=2*n-y1-y2;
        if(x<=t)
            cout<<x+n<<endl;
        else cout<<t+n<<endl;
    }
    else if(q==n)                 //情况2；
    {
        x=x1+x2;
        t=2*n-x1-x2;
        if(x<=t)
            cout<<x+n<<endl;
        else cout<<t+n<<endl;
    }
        else cout<<p+q<<endl;     //情况3；      
    return 0;
}