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
from collections import deque

H, W, K = map(int, input().split())
A = [input() for h in range(H)]

for h in range(H):
  for w in range(W):
    if A[h][w] == 'S':
      sx, sy = h, w

INF = 1000
dist = [[INF for w in range(W)] for h in range(H)]
dist[sx][sy] = 0


dxy = [[1, 0], [0, 1], [-1, 0], [0, -1]]
start = []
d = deque([[sx, sy]])
while d:
  x, y = d.popleft()
  for dx, dy in dxy:
    xx = x + dx; yy = y + dy
    if 0<=xx<H and 0<=yy<W and dist[xx][yy]==INF:
      if A[xx][yy] == '.':
        dist[xx][yy] = dist[x][y] + 1
        d.append([xx, yy])
        if dist[xx][yy] <= K:
          start.append([xx, yy])

ans = min(-(-(H-1-sx)//K)+1, -(-sx//K)+1, -(-(W-1-sy)//K)+1, -(-sy//K)+1)
for x, y in start:
  ans = min(ans, -(-(W-1 - y)//K) + 1) # 右
  ans = min(ans, -(-y//K) + 1) # 左
  ans = min(ans, -(-(H-1 - x)//K) + 1) # 下
  ans = min(ans, -(-x//K) + 1) # 上

print(ans)