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, raw_input().split())
a=["#"*(w+2)]+["#"+raw_input()+"#" for i in range(h)]+["#"*(w+2)]
q=[]
first=0
dis=[[0]*(w+2) for i in range(h+2)]
for i in range(h+2):
	for j in range(w+2):
		if a[i][j]=="S":
			q.append((i, j))
			dis[i][j]=1
while first<len(q):
	cur_x, cur_y=q[first]
	first+=1
	for next_x, next_y in ((cur_x+1, cur_y), (cur_x-1, cur_y), (cur_x, cur_y+1), (cur_x, cur_y-1)):
		if a[next_x][next_y]=="." and not dis[next_x][next_y]:
			q.append((next_x, next_y))
			dis[next_x][next_y]=dis[cur_x][cur_y]+1
result=h*w
for i in range(h+2):
	for j in range(w+2):
		if 0<dis[i][j]<=k+1:
			dis_exit=min(i-1, j-1, h-i, w-j)
			cnt=(dis_exit+k-1)/k
			result=min(result, 1+cnt)
print result