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:
import itertools, math, collections, sys
input = sys.stdin.readline
H, W, K = [int(_) for _ in input().split()]
A = [list(input()) for _ in range(H)]
for i, j in itertools.product(range(H), range(W)):
    if A[i][j] == 'S':
        break
Q = collections.deque([[i, j, 0]])
sH = set()
sW = set()
while Q:
    h, w, dist = Q.popleft()
    if A[h][w] == '#':
        continue
    A[h][w] = '#'
    sH.add(h)
    sW.add(w)
    if dist < K:
        for nh, nw in [[h - 1, w], [h + 1, w], [h, w - 1], [h, w + 1]]:
            if 0 <= nh < H and 0 <= nw < W:
                Q += [[nh, nw, dist + 1]]
print(1 + min([
    math.ceil(_ / K)
    for _ in (min(sH), H - 1 - max(sH), min(sW), W - 1 - max(sW))
]))