Olya loves energy drinks. She loves them so much that her room is full of empty cans from energy drinks.

Formally, her room can be represented as a field of *n* × *m* cells, each cell of which is empty or littered with cans.

Olya drank a lot of energy drink, so now she can run *k* meters per second. Each second she chooses one of the four directions (up, down, left or right) and runs from 1 to *k* meters in this direction. Of course, she can only run through empty cells.

Now Olya needs to get from cell (*x*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">1</sub>) to cell (*x*<sub class="lower-index">2</sub>, *y*<sub class="lower-index">2</sub>). How many seconds will it take her if she moves optimally?

It's guaranteed that cells (*x*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">1</sub>) and (*x*<sub class="lower-index">2</sub>, *y*<sub class="lower-index">2</sub>) are empty. These cells can coincide.
The first line contains three integers *n*, *m* and *k* (1 ≤ *n*, *m*, *k* ≤ 1000) — the sizes of the room and Olya's speed.

Then *n* lines follow containing *m* characters each, the *i*-th of them contains on *j*-th position "#", if the cell (*i*, *j*) is littered with cans, and "." otherwise.

The last line contains four integers *x*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub>, *y*<sub class="lower-index">2</sub> (1 ≤ *x*<sub class="lower-index">1</sub>, *x*<sub class="lower-index">2</sub> ≤ *n*, 1 ≤ *y*<sub class="lower-index">1</sub>, *y*<sub class="lower-index">2</sub> ≤ *m*) — the coordinates of the first and the last cells.
In the first sample Olya should run 3 meters to the right in the first second, 2 meters down in the second second and 3 meters to the left in the third second.

In second sample Olya should run to the right for 3 seconds, then down for 2 seconds and then to the left for 3 seconds.

Olya does not recommend drinking energy drinks and generally believes that this is bad.
3 4 4
....
###.
....
1 1 3 1
3
3 4 1
....
###.
....
1 1 3 1
8
2 2 1
.#
#.
1 1 2 2
-1
10 10 1
##########
#.........
#.#######.
#.#.....#.
#.#.###.#.
#.#.#.#.#.
#.#.#.#.#.
#.#.#...#.
#.#.#####.
#.#.......
6 6 10 2
48
10 10 3
##########
##########
##########
##########
##########
##########
##########
#########.
#########.
####..###.
10 6 10 5
1
10 10 3
...##..#..
#.#..#...#
..#.##.#..
##..#..#.#
..#...##..
.#.#.#....
#......#..
.#.####.##
......#...
.#.##...##
4 6 8 1
7
10 10 1000
..........
..........
..........
..........
..........
..........
..........
..........
..........
..........
7 6 1 3
2
10 10 1000
..........
..........
..........
..........
..........
..........
..........
..........
..........
.....#....
3 9 5 1
2
10 10 10
...#......
##.#.#####
...#...###
.#####.###
...#...###
##.#.#####
...#...###
.#####.###
.......###
##########
1 1 1 10
17
2 5 5
#####
##.##
2 3 2 3
0
4 4 4
...#
.#.#
....
##..
1 1 3 4
2
6 6 100
....##
.##.##
.##.##
......
###.##
###.##
1 1 4 6
2
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")


if os.path.exists('input.txt'):
    sys.stdin = open('input.txt', 'r')
    sys.stdout = open('output.txt', 'w')


dx = [0, 0, -1, 1]
dy = [1, -1, 0, 0]
n, m, k = mp()
l = []
now = [INF for i in range(m)]
distance = []
for i in range(n):
    cur = list(now)
    s = st()
    l.append(s)
    distance.append(cur)
a, b, c, d = mp()
a, b, c, d = a-1, b-1, c-1, d-1
q = deque()
q.append((a, b))
distance[a][b] = 0
v[a][b] = 1
ans = -1
while q:
    x, y = q.popleft()
    if (x, y) == (c, d):
        ans = distance[x][y]
        break
    for i in range(4):
        for j in range(1, k+1):
            x1, x2 = x+j*dx[i], y+j*dy[i]
            if x1 < 0 or x2 < 0 or x1 >= n or x2 >= m or l[x1][x2] == '#' or distance[x1][x2] < 1+distance[x][y]:
                break
            else:
                if distance[x1][x2] > 1 + distance[x][y]:
                    distance[x1][x2] = 1+distance[x][y]
                    q.append((x1, x2))
pr(ans)
