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
h, w, k = [ int(v) for v in input().split() ]
field = [ list(input()) for i in range(h) ]
adjust_list = ((0,1),(0,-1),(1,0),(-1,0))
for i in range(h):
    for j in range(w):
        if field[i][j] == "S":
            s = (i,j)

def nearest(y, x):
    return min(y,h-1-y,x,w-1-x) 


def bfs():
    global field, adjust_list, s
    search_list = [s]
    for _ in range(k):
        new_search_list = []
        for i in search_list:
            for j in adjust_list:
                if 0 <= i[0]+j[0] < h and 0 <= i[1]+j[1] < w:
                    if field[i[0]+j[0]][i[1]+j[1]] == ".":
                        field[i[0]+j[0]][i[1]+j[1]] = "S"
                        new_search_list.append((i[0]+j[0],i[1]+j[1]))
        search_list = new_search_list
    return

bfs()
print(1 + ( min([ nearest(i,j) for j in range(w) for i in range(h) if field[i][j] == "S" ]) + k - 1 ) // k)
