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
import math
h, w, k = [int(item) for item in input().split()]
field = ["O" * (w+2)]
for i in range(h):
    line = "O" + input().rstrip() + "O"
    field.append(line)
    if "S" in line:
        start = [i+1, line.index("S"), 0]
field.append("O" * (w+2))

dq = deque()
dq.append(start)
visited = [[0] * (w+2) for _ in range(h+2)]
visited[start[0]][start[1]] = 1
closest = 10**3 
while dq:
    x, y, d = dq.popleft()
    closest = min(closest, min(x-1, y-1, h-x, w-y))
    if closest == 0:
        break
    d += 1
    for dx, dy in [[1,0], [0,1], [-1,0], [0,-1]]:
        if field[x+dx][y+dy] == "." and visited[x+dx][y+dy] == 0 and d <= k:
            dq.append([x+dx, y+dy, d])
            visited[x+dx][y+dy] = 1
print(1 + int(math.ceil(closest / k)))