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: Accepted)

#include<bits/stdc++.h>

#define rep(i,n) for (int i = 0; i < (n); ++i)

using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

// 定数たち
const ll INF = 1e18;
const double PI = 3.14159265358979323846264338327950;

//最大公約数
ll gcd(ll x, ll y) {
    ll tmp = 0;
    if (x < y){
        tmp=x;
        x=y;
        y=tmp;
    }
    while (y > 0) {
        ll r = x % y;
        x = y;
        y = r;
        }
    return x;
}

//最大公倍数
ll lcm(ll x,ll y){
    return x/gcd(x,y)*y;
}

const int MAX = 1e6+1;
const ll MOD = 1e9+7;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

struct mint {
  ll x; // typedef long long ll;
    mint(ll x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
    }
    mint& operator-=(const mint a) {
    if ((x += MOD-a.x) >= MOD) x -= MOD;
    return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const {
        return pow(MOD-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
};

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

long long modpow(long long a, long long n, long long MOD) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % MOD;
        a = a * a % MOD;
        n >>= 1;
    }
    return res;
}

struct UnionFind {
    vector<int> par;
    UnionFind(int n) : par(n, -1) { }
    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }
    bool issame(int x, int y) {
        return root(x) == root(y);
    }
    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    ll size(int x) {
        return -par[root(x)];
    }
};

// 二項係数（大きい値用）
ll count(int n, int a){
    ll bunshi = 1;
    for(int i = 0; i < a; i++){
        bunshi*=(n-i);
        bunshi%=MOD;
    }
    ll bunbo = 1;
    for(int i = 1; i < a+1; i++){
        bunbo*=i;
        bunbo%=MOD;
    }
    bunbo = modpow(bunbo,MOD-2,MOD);
    return (bunshi*bunbo)%MOD;
}

// 約数列挙。約数をvector<ll>で返す。計算量はsqrt(n)。
vector<ll> divisor(ll n) {
    vector<ll> ret;
    for(ll i = 1; i * i <= n; i++) {
        if(n % i == 0) {
        ret.push_back(i);
        if(i * i != n) ret.push_back(n / i);
        }
    }
    sort(begin(ret), end(ret));
    return (ret);
}


// グラフ
struct edge{ll to, cost;};
// typedef pair<ll,ll> P;
struct graph{
    ll V;
    vector<vector<edge> > G;
    vector<ll> d;

    graph(ll n){
        init(n);
    }

    void init(ll n){
        V = n;
        G.resize(V);
        d.resize(V);
        rep(i,V){
        d[i] = INF;
        }
    }

    void add_edge(ll s, ll t, ll cost){
        edge e;
        e.to = t, e.cost = cost;
        G[s].push_back(e);
    }

    void dijkstra(ll s){
        rep(i,V){
            d[i] = INF;
        }
        d[s] = 0;
        priority_queue<P,vector<P>, greater<P> > que;
        que.push(P(0,s));
        while(!que.empty()){
            P p = que.top(); que.pop();
            ll v = p.second;
            if(d[v]<p.first) continue;
            for(auto e : G[v]){
                if(d[e.to]>d[v]+e.cost){
                d[e.to] = d[v]+e.cost;
                que.push(P(d[e.to],e.to));
                }
            }
        }
    }
};



// ここから開始
const int di[] = {-1,0,1,0};
const int dj[] = {0,-1,0,1};

int main(){
    int h, w;
    cin >> h >> w;
    int si, sj;
    cin >> si >> sj;
    int ti, tj;
    cin >> ti >> tj;
    --si; --sj; --ti; --tj;
    vector<string> s(h);
    rep(i,h) cin >> s[i];
    const int INF = 1e9;
    vector<vector<int> > dist(h,vector<int>(w,INF));
    deque<P> q;
    dist[si][sj] = 0;
    q.emplace_back(si,sj);
    while (!q.empty()) {
        int i = q.front().first;
        int j = q.front().second;
        int d = dist[i][j];
        q.pop_front();
        if (dist[i][j] != d) continue;
        rep(v,4) {
            int ni = i+di[v], nj = j+dj[v];
            if (ni < 0 || ni >= h || nj < 0 || nj >= w) continue;
            if (s[ni][nj] == '#') continue;
            if (dist[ni][nj] <= d) continue;
            dist[ni][nj] = d;
            q.emplace_front(ni,nj);
        }
        for (int ei = -2; ei <= 2; ++ei) {
            for (int ej = -2; ej <= 2; ++ej) {
            int ni = i+ei, nj = j+ej;
            if (ni < 0 || ni >= h || nj < 0 || nj >= w) continue;
            if (s[ni][nj] == '#') continue;
            if (dist[ni][nj] <= d+1) continue;
            dist[ni][nj] = d+1;
            q.emplace_back(ni,nj);
            }
        }
    }
    int ans = dist[ti][tj];
    if (ans == INF) ans = -1;
    cout << ans << endl;
    return 0;
}