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())
m = [input() for i in range(H)]
v = [[0]*W for i in range(H)]
f = 0
for i in range(H):
	if f:
		break
	for j in range(W):
		if m[i][j] == "S":
			x = i
			y = j
			f = 1
			break
dq = deque([[x,y,0]])
v[x][y] = 1
ans = (min(x, y, H-1-x, W-1-y)+K-1)//K
while dq:
	x, y, c = dq.popleft()
	ans = min(ans, (min(x, y, H-1-x, W-1-y)+K-1)//K)
	if c < K:
		c += 1
		if x > 0 and m[x-1][y] == "." and not v[x-1][y]:
			dq.append([x-1,y,c])
			v[x-1][y] = 1
		if x < H-1 and m[x+1][y] == "." and not v[x+1][y]:
			dq.append([x+1,y,c])
			v[x+1][y] = 1
		if y > 0 and m[x][y-1] == "." and not v[x][y-1]:
			dq.append([x,y-1,c])
			v[x][y-1] = 1
		if y < W-1 and m[x][y+1] == "." and not v[x][y+1]:
			dq.append([x,y+1,c])
			v[x][y+1] = 1
print(ans+1)