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
h,w,k=map(int,input().split())
b=[list("#"*(w+2))]+[list("#"+input()+"#")for _ in range(h)]+[list("#"*(w+2))]
for i in range(h+2):
  for j in range(w+2):
    if b[i][j]=="S":
      Q=[(i,j)]
      ans=0--min(i-1,h-i,j-1,w-j)//k
co=0
dx=[1,0,-1,0]
dy=[0,1,0,-1]
while Q and co!=k+1:
  co+=1
  P=[]
  for i,j in Q:
    ans=min(ans,0--min(i-1,h-i,j-1,w-j)//k)
    for x,y in zip(dx,dy):
      if b[i+x][j+y]==".":
        b[i+x][j+y]=0
        P.append((i+x,j+y))
        if i+x in [1,h] or j+y in [1,w]:print(1);exit()
  Q=P
print(ans+1)