#include<bits/stdc++.h>
#define MAX 200

using namespace std;

int dx[] = {-1, 1, 2, 1,-1,-2};       // standard always follow.
int dy[] = { 2, 2, 0,-2,-2, 0};       // standard always follow.
string path[]={"UL","UR","R","LR","LL","L"};
int dis[MAX][MAX][6];           // editorial sol.

char s[MAX][MAX]={'0'};
int n;

struct node {                       // learn to make nodes.
    int x, y, dir;
    node() { }                      // an empty node creation.
    node(int x, int y, int dir) {   // x , t denote the current position and the dir 
                                    // ditermines what ???
        this->x = x, this->y = y, this->dir = dir;
    }
};

bool valid(int x, int y)            // I did this in the first implementation.
{
    if ( x < 0 || y < 0 || x >= n || y >= n ) return false;
    //if ( s[x][y] == '*' ) return false;
    return true;
    
}

int main() {
    cin>>n;
    int sx,sy,ex,ey;
    cin>>sx>>sy>>ex>>ey;
    int cnt1 = 0, cnt2 = 0;
    
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            for ( int k = 0; k < 6; k++ ) dis[i][j][k] = 0;
            if(i==sx && j==sy){
                s[i][j]='V';
                cnt1++;
            }
            else if(i==ex && j==ey){
                s[i][j]='H';
                cnt2++;
            }
            else{
                s[i][j]='0';
            }
        }
    }
    
    int init = (ex - sx)*(ey - sy);
    
    queue <node> q;
    
    for ( int k = 0; k < 6; k++ ) {
        dis[sx][sy][k] = 0;
        q.push(node(sx, sy, k));
    }
    
    int ans=0;
    
    while ( !q.empty() ) {
        node f = q.front();
        q.pop();
        for ( int k = 0; k < 6; k++ ) {         // bfs traversal in a grid.
            int new_x = f.x + dx[k];            // thus making use of dx and dy.
            int new_y = f.y + dy[k];
            if ( valid(new_x, new_y) && ((new_x - f.x)*(new_x - f.x) + (new_y - f.y)*(new_y - f.y)) < init ) {
                    dis[new_x][new_y][k] = dis[f.x][f.y][f.dir] + 1;
                    q.push(node(new_x, new_y, k));
            }
            if(new_x == ex && new_y == ey){
                ans = dis[new_x][new_y][k];
                break;
            }
        }
    }
    
    printf("%d\n", ans);
    
    return 0;
}