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:
def main(h,w,k,a):
  s=-1,-1
  for i in range(h):
    for j in range(w):
      if a[i][j]=='S':
        s=i,j
        break
    if s[0]!=-1:break
  inf=float('inf')
  da=[[inf]*w for _ in range(h)]
  da[s[0]][s[1]]=0
  todo=[s]
  dst=min(s[0],h-s[0]-1,s[1],w-s[1]-1)
  cnt=0
  while cnt<k and todo:
    ntodo=[]
    while todo:
      i,j=todo.pop()
      for di,dj in ((1,0),(-1,0),(0,1),(0,-1)):
        if 0<=i+di<h and 0<=j+dj<w and a[i+di][j+dj]=='.' and da[i+di][j+dj]>da[i][j]+1:
          da[i+di][j+dj]=da[i][j]+1
          ntodo.append((i+di,j+dj))
          dst=min(dst,i+di,h-i-di-1,j+dj,w-j-dj-1)
    cnt+=1
    todo=ntodo
  return 1+(dst+k-1)//k

h,w,k=map(int,input().split())
a=[input() for _ in range(h)]
print(main(h,w,k,a))