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, raw_input().split())

MAP = [raw_input() for _ in xrange(H)]
visited = [[False for _ in xrange(W)] for _ in xrange(H)]

for y in range(H):
    x = MAP[y].find('S')
    if x != -1:
        sx = x
        sy = y

Q = deque()
Q.append((0, (sx, sy)))
visited[sy][sx] = True
while len(Q) > 0:
    cost, [x, y] = Q.popleft()
    for dx, dy in [(-1, 0), (1, 0), (0, 1), (0, -1)]:
        px, py = x+dx, y+dy
        if 0 <= px < W and 0 <= py < H:
            if not visited[py][px] and MAP[py][px] != "#":
                visited[py][px] = True
                if cost + 1 < K:
                    Q.append((cost + 1, (px, py)))

d = 10000
for x in xrange(W):
    for y in xrange(H):
        if visited[y][x]:
            d = min(d, x, W-1-x)
            d = min(d, y, H-1-y)

print (d + K - 1) / K + 1
