Problem: Takahashi is locked within a building.

This building consists of H×W rooms, arranged in H rows and W columns. We will denote the room at the i-th row and j-th column as (i,j). The state of this room is represented by a character A_{i,j}. If A_{i,j}= `#`, the room is locked and cannot be entered; if A_{i,j}= `.`, the room is not locked and can be freely entered. Takahashi is currently at the room where A_{i,j}= `S`, which can also be freely entered.

Each room in the 1-st row, 1-st column, H-th row or W-th column, has an exit. Each of the other rooms (i,j) is connected to four rooms: (i-1,j), (i+1,j), (i,j-1) and (i,j+1).

Takahashi will use his magic to get out of the building. In one cast, he can do the following:

* Move to an adjacent room at most K times, possibly zero. Here, locked rooms cannot be entered.
* Then, select and unlock at most K locked rooms, possibly zero. Those rooms will remain unlocked from then on.



His objective is to reach a room with an exit. Find the minimum necessary number of casts to do so.

It is guaranteed that Takahashi is initially at a room without an exit.

Constraints

* 3 ≤ H ≤ 800
* 3 ≤ W ≤ 800
* 1 ≤ K ≤ H×W
* Each A_{i,j} is `#` , `.` or `S`.
* There uniquely exists (i,j) such that A_{i,j}= `S`, and it satisfies 2 ≤ i ≤ H-1 and 2 ≤ j ≤ W-1.

Input

Input is given from Standard Input in the following format:


H W K
A_{1,1}A_{1,2}...A_{1,W}
:
A_{H,1}A_{H,2}...A_{H,W}


Output

Print the minimum necessary number of casts.

Examples

Input

3 3 3
#.#
#S.
###


Output

1


Input

3 3 3
.#
S.


Output

1


Input

3 3 3

S#


Output

2


Input

7 7 2


...##
S###
.#.##
.###


Output

2

Solution (in PYTHON3):

Solution:
from collections import deque
h, w, k = map(int, input().split())
maze = [input() for _ in range(h)]
visited = [[-1]*w for j in range(h)]
for i in range(h):
    if 'S' in maze[i]:
        sy, sx = i, maze[i].index('S')
queue = deque([[sy, sx]])
visited[sy][sx] = 0
r = 10**18
while queue:
    y, x = queue.popleft()
    r = min(r, y, x, h-1-y, w-1-x)
    if visited[y][x] == k: continue
    for i, j in ([1, 0], [-1, 0], [0, 1], [0, -1]):
        new_y, new_x = y+i, x+j
        if new_y < 0 or new_y >= h: continue
        if new_x < 0 or new_x >= w: continue
        if maze[new_y][new_x] == '.' and visited[new_y][new_x] == -1:
            visited[new_y][new_x] = visited[y][x] + 1
            queue.append([new_y, new_x])
print(1 -(-r // k))