Problem:
Score : 600 points

Problem Statement

Snuke, a water strider, lives in a rectangular pond that can be seen as a grid with H east-west rows and W north-south columns. Let (i,j) be the square at the i-th row from the north and j-th column from the west.

Some of the squares have a lotus leaf on it and cannot be entered.
The square (i,j) has a lotus leaf on it if c_{ij} is @, and it does not if c_{ij} is ..

In one stroke, Snuke can move between 1 and K squares (inclusive) toward one of the four directions: north, east, south, and west.
The move may not pass through a square with a lotus leaf. Moving to such a square or out of the pond is also forbidden.

Find the minimum number of strokes Snuke takes to travel from the square (x_1,y_1) to (x_2,y_2).
If the travel from (x_1,y_1) to (x_2,y_2) is impossible, point out that fact.

Constraints

1 \leq H,W,K \leq 10^6

H \times W \leq 10^6

1 \leq x_1,x_2 \leq H

1 \leq y_1,y_2 \leq W

x_1 \neq x_2 or y_1 \neq y_2.

c_{i,j} is . or @.

c_{x_1,y_1} = .

c_{x_2,y_2} = .

All numbers in input are integers.

Input

Input is given from Standard Input in the following format:

H W K
x_1 y_1 x_2 y_2
c_{1,1}c_{1,2} .. c_{1,W}
c_{2,1}c_{2,2} .. c_{2,W}
:
c_{H,1}c_{H,2} .. c_{H,W}

Output

Print the minimum number of strokes Snuke takes to travel from the square (x_1,y_1) to (x_2,y_2), or print -1 if the travel is impossible.

Sample Input 1

3 5 2
3 2 3 4
.....
.@..@
..@..

Sample Output 1

5

Initially, Snuke is at the square (3,2).
He can reach the square (3, 4) by making five strokes as follows:

From (3, 2), go west one square to (3, 1).

From (3, 1), go north two squares to (1, 1).

From (1, 1), go east two squares to (1, 3).

From (1, 3), go east one square to (1, 4).

From (1, 4), go south two squares to (3, 4).

Sample Input 2

1 6 4
1 1 1 6
......

Sample Output 2

2

Sample Input 3

3 3 1
2 1 2 3
.@.
.@.
.@.

Sample Output 3

-1

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using P = pair<ll, ll>;
using Pld = pair<ld, ld>;
using Vec = vector<ll>;
using VecP = vector<P>;
using VecB = vector<bool>;
using VecC = vector<char>;
using VecD = vector<ld>;
using VecS = vector<string>;
template <typename T>
using Vec2 = vector<vector<T>>;
#define REP(i, m, n) for(ll i = (m); i < (n); ++i)
#define REPN(i, m, n) for(ll i = (m); i <= (n); ++i)
#define REPR(i, m, n) for(ll i = (m)-1; i >= (n); --i)
#define REPNR(i, m, n) for(ll i = (m); i >= (n); --i)
#define rep(i, n) REP(i, 0, n)
#define repn(i, n) REPN(i, 1, n)
#define repr(i, n) REPR(i, n, 0)
#define repnr(i, n) REPNR(i, n, 1)
#define all(s) (s).begin(), (s).end()
#define pb push_back
#define fs first
#define sc second
template <typename T>
bool chmax(T &a, const T b){if(a < b){a = b; return true;} return false;}
template <typename T>
bool chmin(T &a, const T b){if(a > b){a = b; return true;} return false;}
template <typename T>
ll pow2(const T n){return (1LL << n);}
template <typename T>
void cosp(const T n){cout << n << ' ';}
void co(void){cout << '\n';}
template <typename T>
void co(const T n){cout << n << '\n';}
template <typename T1, typename T2>
void co(pair<T1, T2> p){cout << p.fs << ' ' << p.sc << '\n';}
template <typename T>
void co(const vector<T> &v){for(T i : v) cosp(i); co();}
template <typename T>
void co(initializer_list<T> v){for(T i : v) cosp(i); co();}
void ce(void){cerr << '\n';}
template <typename T>
void ce(const T n){cerr << n << '\n';}
template <typename T>
void cesp(const T n){cerr << n << ' ';}
template <typename T>
void ce(initializer_list<T> v){for(T i : v) cesp(i); ce();}
void sonic(){ios::sync_with_stdio(false); cin.tie(0);}
void setp(const ll n){cout << fixed << setprecision(n);}
constexpr int INF = 1e9+1;
constexpr ll LINF = 1e18L+1;
constexpr ll MOD = 1e9L+7;
// constexpr ll MOD = 998244353;
constexpr ld EPS = 1e-11;
const double PI = acos(-1);

ll h, w, k;
VecS board;
Vec2<ll> dist;
ll dx[4] = {1, -1, 0, 0};
ll dy[4] = {0, 0, 1, -1};

bool in_board(ll x, ll y){
	return x >= 0 && x < h && y >= 0 && y < w && board[x][y] == '.';
}

void bfs(ll fx, ll fy){
	dist[fx][fy] = 0;
	queue<P> que;
	que.push({fx, fy});
	while(!que.empty()){
		P p = que.front(); que.pop();
		// co(p);
		ll a = p.first, b = p.second;
		rep(i, 4){
			ll x = a, y = b;
			repn(j, k){
				x += dx[i], y += dy[i];
				// co({x, y});
				if(in_board(x, y)){
					if(chmin(dist[x][y], dist[a][b] + 1)){
						que.push({x, y});
					}else if(dist[x][y] == dist[a][b] + 1){
						continue;
					} else break;
				} else break;
			}
		}
	}
}

int main(void){
	cin >> h >> w >> k;
	dist.resize(h);
	rep(i, h) dist[i].resize(w, LINF);
	ll x1, y1, x2, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	--x1, --y1, --x2, --y2;
	board.resize(h);
	rep(i, h) cin >> board[i];

	bfs(x1, y1);

	ll ans = dist[x2][y2];
	/*
	rep(i, h){
		rep(j, w) cosp(dist[i][j]);
		co();
	}
	*/
	if(ans == LINF) ans = -1;
	co(ans);

	return 0;
}