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
import math
H,W,K = map(int,input().split())
HW = [list(input()) for i in range(H)]
for i,h in enumerate(HW):
	if "S" in h:
		y = i
		x = h.index("S")
A = [(x,y)]
m = 800
def idou(a,b):
	X = []
	for i2 in range(4):
		if i2 == 0:
			t,k = a - 1, b
		elif i2 == 1:
			t,k = a, b - 1
		elif i2 == 2:
			t,k = a + 1, b
		elif i2 == 3:
			t,k = a, b + 1
		if 0 <= t < W and 0 <= k < H and HW[k][t] == ".":
			HW[k][t] = "o"
			X.append((t,k))
	return X
cnt = 0
while cnt < K:
	cnt += 1
	B = []
	for p in A:
		x,y = p
		b = idou(x,y)
		B.extend(b)
	A = B[:]
M = []
for i,w in enumerate(zip(*HW)):
	#print(w)
	if "o" in w or "S" in w:
		M.append(min(i,W - i - 1))
		
for i,h in enumerate(HW):
	if "S" in h or "o" in h:
		M.append(min(i, H - i - 1))
		#print(h)
m = min(M)
print((m + K - 1) // K + 1)