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
from math import ceil
H,W,K=map(int,input().split())
grid=[input() for i in range(H)]

for i in range(H):
  for j in range(W):
    if grid[i][j]=="S":
      S=(i,j)

stack=deque([(S[0],S[1],0)])
visited=[[False]*(W) for i in range(H)]
visited[S[0]][S[1]]=True
a=float("inf")
t=[-1,1]
while stack:
  i,j,m=stack.popleft()
  if m>K:
    break
  a=min(a,min(i,H-1-i,j,W-1-j))
  if a==0:
    break
  for s in t:
    if not visited[i+s][j] and grid[i+s][j]!="#":
      stack.append((i+s,j,m+1))
    if not visited[i][j+s] and grid[i][j+s]!="#":
      stack.append((i,j+s,m+1))
    visited[i+s][j]=True
    visited[i][j+s]=True
ans=1+ceil(a/K)
print(ans)