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:
from collections import deque

INF = float("inf")

H, W, K = map(int, input().split())
A = [list(input()) for _ in range(H)]

Si, Sj = (0, 0)
for i, a in enumerate(A):
    if "S" in a:
        Si, Sj = (i, a.index("S"))
        A[Si][Sj] = "c"
        break

distance = INF

Q = deque([(0, Si, Sj)])
while Q:
    cnt, i, j = Q.popleft()

    distance = min(distance, i, j, H - i - 1, W - j - 1)

    if cnt >= K:
        continue

    for ni, nj in ((i + 1, j), (i, j + 1), (i - 1, j), (i, j - 1)):
        if 0 <= ni < H and 0 <= nj < W and A[ni][nj] == ".":
            A[ni][nj] = "c"
            Q.append((cnt + 1, ni, nj))

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