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 queue import deque
from math import ceil

h, w, k = map(int, input().split())

rooms = [None] * h
for i in range(h):
    row = input().strip()
    rooms[i] = row
    if 'S' in row:
        s = (i, row.index('S'))


def generate_neighbors(i, j):
    for ni, nj in [(i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1)]:
        if rooms[ni][nj] == '.':
            yield ni, nj


def get_dist_to_exit(i, j):
    return min(i, h - i - 1, j, w - j - 1)


def first_bws():
    visited = {s}
    queue = deque((1, ij) for ij in generate_neighbors(*s))
    min_exit = get_dist_to_exit(*s)
    while queue:
        dist, ij = queue.popleft()
        if dist > k:
            break
        if ij in visited:
            continue
        i, j = ij
        tmp_min_exit = get_dist_to_exit(i, j)
        if tmp_min_exit == 0:
            return 0
        min_exit = min(min_exit, tmp_min_exit)
        visited.add(ij)
        queue.extend((dist + 1, nij) for nij in generate_neighbors(i, j) if nij not in visited)
    return min_exit


dist_to_exit = first_bws()
print(ceil(dist_to_exit / k) + 1)