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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

#define INF 100000000
#define YJ 1145141919
#define INF_INT_MAX 2147483647
#define INF_LL 9223372036854775
#define INF_LL_MAX 9223372036854775807
#define EPS 1e-10
#define MOD 1000000007
#define MOD9 998244353
#define Pi acos(-1)
#define LL long long
#define ULL unsigned long long
#define LD long double

#define int long long

using II = pair<int, int>;

int gcd(int a, int b) { return b != 0 ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
int extgcd(int a, int b, int &x, int &y) { int g = a; x = 1; y = 0; if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x; return g; }

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(a)  begin((a)), end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())

int H, W, K;
vector<string> field;
int sx, sy, gx, gy;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};

bool isInField(int x, int y) {
  return 0 <= x && x < W && 0 <= y && y < H;
}

struct Status {
  Status(int x, int y, int turn) : x(x), y(y), turn(turn) {}
  int x, y, turn;
};

int solve() {
  unordered_map<int, unordered_map<int, bool>> used;
  unordered_map<int, unordered_map<int, bool>> preGoal;

  REP(d,4) {
    FOR(k,1,K+1) {
      int nx = gx + k * dx[d];
      int ny = gy + k * dy[d];
      if (!isInField(nx, ny)) {
        break;
      }
      if (field[ny][nx] == '@') {
        break;
      }
      preGoal[ny][nx] = true;
    }
  }

  queue<Status> que;
  que.push(Status(sx, sy, 0));
  used[sy][sx] = true;

  while(que.size()) {
    Status status = que.front(); que.pop();

    if (preGoal[status.y][status.x]) {
      return status.turn+1;
    }

    REP(d,4) {
      FOR(k,1,K+1) {
        int nx = status.x + k*dx[d];
        int ny = status.y + k*dy[d];
        if (!isInField(nx, ny)) {
          break;
        }
        if (used[ny][nx]) {
          break;
        }
        if (field[ny][nx] == '@') {
          break;
        }
        used[ny][nx] = true;
        que.push(Status(nx, ny, status.turn+1));
      }
    }
  }

  return -1;
}

signed main()
{
  cin >> H >> W >> K;
  cin >> sy >> sx >> gy >> gx;
  sy--; sx--; gy--; gx--;

  field.resize(H);
  REP(h,H) {
    cin >> field[h];
  }

  cout << solve() << endl;

  return 0;
}