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:

h,w,k=map(int,input().split())
maze=[]

for i in range(h):
    maze.append(input())
    for j in range(w):
        if maze[i][j]=="S":x,y=i,j
    
ans=float("INF")
l=[[float("INF")]*w for i in range(h)]
l[x][y]=0
from collections import deque
dq=deque([(x,y)])
#pop/append/(append,pop)_left/in/len/count/[]/index/rotate()(右へnずらす)
p={(x,y)}
while len(dq):
    nx,ny=dq.popleft()
    for a,s in [(0,1),(0,-1),(1,0),(-1,0)]:
        if 0<=a+nx<h and 0<=ny+s<w and l[a+nx][s+ny]==float("INf")and maze[a+nx][s+ny]==".":
            l[a+nx][ny+s]=l[nx][ny]+1
            if l[a+nx][ny+s]<=k:p.add((a+nx,s+ny))
            if l[a+nx][ny+s]<k:dq.append((a+nx,s+ny))
ans=float("INf")

for x,y in p:
    ans=min(ans,(min(x,y,h-1-x,w-1-y)+k-1)//k+1)
print(ans)