Problem:
Score : 400 points

Problem Statement

A maze is composed of a grid of H \times W squares - H vertical, W horizontal.

The square at the i-th row from the top and the j-th column from the left - (i,j) - is a wall if S_{ij} is # and a road if S_{ij} is ..

There is a magician in (C_h,C_w). He can do the following two kinds of moves:

Move A: Walk to a road square that is vertically or horizontally adjacent to the square he is currently in.

Move B: Use magic to warp himself to a road square in the 5\times 5 area centered at the square he is currently in.

In either case, he cannot go out of the maze.

At least how many times does he need to use the magic to reach (D_h, D_w)?

Constraints

1 \leq H,W \leq 10^3

1 \leq C_h,D_h \leq H

1 \leq C_w,D_w \leq W

S_{ij} is # or ..

S_{C_h C_w} and S_{D_h D_w} are ..

(C_h,C_w) \neq (D_h,D_w)

Input

Input is given from Standard Input in the following format:

H W
C_h C_w
D_h D_w
S_{11}\ldots S_{1W}
\vdots
S_{H1}\ldots S_{HW}

Output

Print the minimum number of times the magician needs to use the magic. If he cannot reach (D_h,D_w), print -1 instead.

Sample Input 1

4 4
1 1
4 4
..#.
..#.
.#..
.#..

Sample Output 1

1

For example, by walking to (2,2) and then using the magic to travel to (4,4), just one use of magic is enough.

Note that he cannot walk diagonally.

Sample Input 2

4 4
1 4
4 1
.##.
####
####
.##.

Sample Output 2

-1

He cannot move from there.

Sample Input 3

4 4
2 2
3 3
....
....
....
....

Sample Output 3

0

No use of magic is needed.

Sample Input 4

4 5
1 2
2 5
#.###
####.
#..##
#..##

Sample Output 4

2

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define irep(i,n) for(int i = 0; i < (int)(n); i++)
#define irep2(i,a,n) for(int i = (int)(a); i <= (int)(n); i++)
#define irrep(i,n) for(int i = (int)(n-1); i > -1; i--)
#define irrep2(i,n,a) for(int i = (int)(n); i >= (int)(a); i--)

#define fi first
#define se second

using namespace std;

using ll = long long;
using v_int = vector<int>;
using v2_int = vector<v_int>;
using v_ll = vector<ll>;
using v2_ll = vector<v_ll>;
using v_string = vector<string>;
using v_bool = vector<bool>;
using v2_bool = vector<v_bool>;
using pii = pair<int, int>;
using mii = map<int, int>;

const double PI = 3.1415926535897932;
const int INF = (int)2e9;
const ll LINF = (ll)2e18;
const ll MOD = 1000000007;
const int dc[8] = {1, -1, 0,  0, 1, -1,  1, -1};
const int dr[8] = {0,  0, 1, -1, 1,  1, -1, -1};

template<typename T, typename U> inline
ostream& operator<<(ostream &o, pair<T,U> &p) {
    return o << "{" << p.first << "," << p.second << "}";
}
template<typename T> inline
istream& operator>>(istream &is, vector<T> &vec) {
    for(auto &v : vec) { is >> v; }
    return is;
}
template<typename T> inline
ostream& operator<<(ostream &os, vector<T> &vec) {
    for(auto &v : vec) { os << v << ",";}
    return os;
}
template<typename T> inline
ostream& operator<<(ostream &os, vector<vector<T> > &mat) {
    for(auto &row : mat) { os << row << endl; }
    return os;
}
template<typename T> inline
void chmin(T &a, T b) { a = min(a, b); }
template<typename T> inline
void chmax(T &a, T b) { a = max(a, b); }

int main(void)
{
    int h, w;
    cin >> h >> w;
    int cr, cl, gr, gl;
    cin >> cr >> cl >> gr >> gl;
    cr--;
    cl--;
    gr--;
    gl--;

    v_string s(h);
    cin >> s;

    v2_int dist(h, v_int(w,INF));
    
    queue<pii> q;
    
    q.push({cr,cl});
    dist[cr][cl] = 0;

    int ans = 0;

    while(q.size()) {
        pii now = q.front();
        q.pop();
        
        bool ng = true;
        irep(i, 4) {
            int row = now.fi + dr[i];
            int col = now.se + dc[i];
            if(row < 0 || row >= h) { continue; }
            if(col < 0 || col >= w) { continue; }
            if(dist[row][col] != INF) { continue; }
            if(s[row][col] == '#') { continue; }
            ng = false;
            dist[row][col] = dist[now.fi][now.se];
            q.push({row,col});
        }
        if(ng) {
            for(int i = -2; i <= 2; i++) {
                for(int j = -2; j <= 2; j++) {
                    int row = now.fi + i;
                    int col = now.se + j;
                    if(row < 0 || row >= h) { continue; }
                    if(col < 0 || col >= w) { continue; }
                    if(dist[row][col] != INF) { continue; }
                    if(s[row][col] == '#') { continue; }
                    dist[row][col] = min(dist[row][col], dist[now.fi][now.se] + 1);
                    q.push({row,col});
                }
            }
        }
    }
    if(dist[gr][gl] == INF) {
        cout << -1 << endl;
    } else {
        cout << dist[gr][gl] << endl;
    }
    return 0;
} /*atcoder*/