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 <algorithm>
#include <bitset>
#include <climits>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

#define MOD 1000000007
#define MOD2 998244353
#define int long long
#define double long double
#define EPS 1e-9
//#define PI 3.14159265358979

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

using namespace std;

template < typename T >
ostream &operator<<(ostream &os, const vector< T > &A) {
	for (int i = 0; i < A.size(); i++)
		os << A[i] << " ";
	os << endl;
	return os;
}
template <>
ostream &operator<<(ostream &os, const vector< vector< int > > &A) {
	int N = A.size();
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < A[i].size(); j++)
			os << A[i][j] << " ";
		os << endl;
	}
	return os;
}

typedef pair< int, int > pii;
typedef long long ll;

struct edge {
	int from, to, d, c;
	edge(int _from = 0, int _to = 0, int _d = 0, int _c = 0) {
		from = _from;
		to = _to;
		d = _d;
		c = _c;
	}
	bool operator<(const edge &rhs) const {
		return (d == rhs.d) ? (c < rhs.c) : (d < rhs.d);
	}
};
struct aabb {
	int x1, y1, x2, y2;
	aabb(int x1, int y1, int x2, int y2) : x1(x1), y1(y1), x2(x2), y2(y2) {}
};

typedef vector< edge > edges;
typedef vector< edges > graph;
struct flow {
	int to, cap, rev, cost;
	flow(int to = 0, int cap = 0, int rev = 0, int cost = 0) : to(to), cap(cap), rev(rev), cost(cost) {}
};
typedef vector< vector< flow > > flows;

const int di[4] = {0, -1, 0, 1};
const int dj[4] = {-1, 0, 1, 0};
const int ci[5] = {0, 0, -1, 0, 1};
const int cj[5] = {0, -1, 0, 1, 0};
const ll LINF = LLONG_MAX / 2;
const int INF = INT_MAX / 2;
const double PI = acos(-1);

int pow2(int n) { return 1LL << n; }
template < typename T, typename U >
bool chmin(T &x, const U &y) {
	if (x > y) {
		x = y;
		return true;
	}
	return false;
}
template < typename T, typename U >
bool chmax(T &x, const U &y) {
	if (x < y) {
		x = y;
		return true;
	}
	return false;
}

struct initializer {
	initializer() {
		cout << fixed << setprecision(20);
	}
};
initializer _____;
vector< int > dijkstra(graph &G, int s) {
	///stat must have "<"&">"
	///init d=vector<stat>
	vector< int > d(G.size(), LINF);
	d[s] = 0;
	//init q=p-queue<pair<stat,index> >
	priority_queue< pii, vector< pii >, greater< pii > > q;
	q.push(pii(0, s));

	while (!q.empty()) {
		pii p = q.top();
		q.pop();
		if (d[p.second] < p.first)
			continue;
		for (int i = 0; i < G[p.second].size(); i++) {
			edge e = G[p.second][i];
			if (p.first + e.d < d[e.to]) {
				d[e.to] = p.first + e.d;
				q.push(pii(d[e.to], e.to));
			}
		}
	}
	return d;
}
int N, M, K, T, Q, H, W;
signed main() {
	cin >> H >> W >> K;
	int x1, y1, x2, y2;
	cin >> x1 >> y1 >> x2 >> y2;
	--x1;
	--y1;
	--x2;
	--y2;
	vector< string > C(H);
	rep(i, H) cin >> C[i];
	vector< vector< int > > grid(H, vector< int >(W, 0));
	rep(i, H) rep(j, W) if (C[i][j] == '@') grid[i][j] = 1;
	vector< vector< vector< pii > > > d(H, vector< vector< pii > >(W, vector< pii >(4, pii(LINF, LINF))));
	d[x1][y1][0] = pii(0, 0);
	priority_queue< pair< pii, pair< pii, int > >, vector< pair< pii, pair< pii, int > > >, greater< pair< pii, pair< pii, int > > > > q;
	pair< pii, pair< pii, int > > s;
	s.first = pii(0, 0);
	s.second.first = pii(x1, y1);
	s.second.second = 0;
	q.push(s);
	while (!q.empty()) {
		pair< pii, pair< pii, int > > p = q.top();
		q.pop();
		int i = p.second.first.first;
		int j = p.second.first.second;
		int k = p.second.second;
		pii dist = p.first;
		if (d[i][j][k] < dist)
			continue;
		int ni = i + di[k];
		int nj = j + dj[k];
		if (0 <= ni && ni < H && 0 <= nj && nj < W && (!grid[ni][nj])) {
			pii ndist = dist;
			ndist.second++;
			if (ndist.second >= K) {
				ndist.second -= K;
				ndist.first++;
			}
			if (ndist < d[ni][nj][k]) {
				d[ni][nj][k] = ndist;
				q.push(make_pair(ndist, pair< pii, int >(pii(ni, nj), k)));
			}
		}
		int nk = k;
		pii ndist = dist;
		if (ndist.second > 0) {
			ndist.first++;
			ndist.second = 0;
		}
		rep(_, 3) {
			nk = (nk + 1) % 4;
			if (ndist < d[i][j][nk]) {
				d[i][j][nk] = ndist;
				q.push(make_pair(ndist, pair< pii, int >(pii(i, j), nk)));
			}
		}
	}
	pii ans = pii(LINF, LINF);
	rep(k, 4) {
		chmin(ans, d[x2][y2][k]);
	}
	if (ans == pii(LINF, LINF)) {
		cout << -1 << endl;
	} else {
		if (ans.second > 0)
			ans.first++;
		cout << ans.first << endl;
	}
	return 0;
}