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())
A = [raw_input() for i in xrange(h)]
sx = sy = 0
for i in xrange(h):
    idx = A[i].find('S')
    if idx != -1:
        sx = idx; sy = i
        break

deq = deque()
deq.append((0, (sx, sy)))

used = set([(sx, sy)])
dd = [[-1, 0], [0, -1], [1, 0], [0, 1]]

while deq:
    cost, [x, y] = deq.popleft()
    for dx, dy in dd:
        nx = x + dx; ny = y + dy
        if 0 <= nx < w and 0 <= ny < h:
            nkey = (nx, ny)
            if A[ny][nx] != '#' and nkey not in used:
                used.add(nkey)
                if cost + 1 < k:
                    deq.append((cost+1, nkey))
ans = 10**18
for x, y in used:
    dx = min(x, w-1-x)
    dy = min(y, h-1-y)
    ans = min(ans, min(dx, dy))
print (ans+k-1)/k + 1
