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=["" for i in range(H)]
for i in range(H):
    A[i]=input()

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

move=[(0,1),(1,0),(0,-1),(-1,0)]
start=set([s])
que=deque([(0,s[0],s[1])])
while que:
    count,posx,posy=que.popleft()
    if count==K:
        continue

    for i,j in move:
        nx=posx+i;ny=posy+j
        if H-1>=nx>=0 and W-1>=ny>=0 and A[nx][ny]!="#" and (nx,ny) not in start:
            start.add((nx,ny))
            que.append((count+1,nx,ny))


ans=float("inf")
for x,y in start:
    test=1+min((abs(x-0)-1)//K+1,(abs(H-1-x)-1)//K+1,(abs(y-0)-1)//K+1,(abs(W-1-y)-1)//K+1)
    ans=min(ans,test)

if s[0]==0 or s[0]==H-1 or s[1]==0 or s[1]==W-1:
    ans=0
print(ans)