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
import sys
input = sys.stdin.readline
from collections import deque

h, w, k = map(int, input().split())
A = [input().rstrip() for _ in range(h)]
for i, a in enumerate(A):
  if "S" in a:
    sy = i
    sx = a.index("S")
    break

que = deque()
D = ((-1, 0), (0, -1), (1, 0), (0, 1))
        
def bfs(y, x):
  seen = [0]*h
  seen[y] |= 1 << x
  F = deque()
  F.append((y, x, 0))
  while F:
    ny, nx, cnt = F.popleft()
    que.append((ny, nx, k))
    if cnt == k:
      continue
    for dy, dx in D:
      yy, xx = ny+dy, nx+dx
      if 0 <= yy < h and 0 <= xx < w:
        if A[yy][xx] == "#":
          continue
        if seen[yy] >> xx & 1:
          continue
        seen[yy] |= 1 << xx
        F.append((yy, xx, cnt+1))

bfs(sy, sx)

seen = [0]*h
while que:
  y, x, cnt = que.popleft()
  if y == 0 or y == h-1 or x == 0 or x == w-1:
    ans = -(-cnt // k)
    break
  for dy, dx in D:
    ny, nx = y+dy, x+dx
    if 0 <= ny < h and 0 <= nx < w:
      if seen[ny] >> nx & 1:
        continue
      seen[ny] |= 1 << nx
      que.append((ny, nx, cnt+1))

print(ans)