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
inf = 10 ** 9 + 7
h, w, k = map(int, input().split())
a = [list(input()) for i in range(h)]
for i in range(h):
    for j in range(w):
        if a[i][j] == "S":
            sx, sy = i, j
d = [(1, 0), (-1, 0), (0, 1), (0, -1)]
l = [[inf] * w for i in range(h)]
l[sx][sy] = 0
s = set()
s.add((sx, sy))
que = deque([[sx, sy]])
ans = inf
while que:
    x, y = que.popleft()
    if x == 0 or x == h - 1 or y == 0 or y == w - 1:
        ans = 1
        break
    if l[x][y] == k:
        continue
    for i in range(4):
        nx, ny = x + d[i][0], y + d[i][1]
        if nx < 0 or nx >= h or ny < 0 or ny >= w or a[nx][ny] == "#" or l[nx][ny] != inf:
            continue
        s.add((nx, ny))
        l[nx][ny] = l[x][y] + 1
        que.append([nx, ny])
for i in s:
    c = (min(i[0], h - i[0] - 1, i[1], w - i[1] - 1) + k - 1) // k + 1
    ans = min(ans, c)
print(ans)