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

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

sx, sy = 0, 0
for y in range(H) :
    flag = False
    for x in range(W) :
        if A[y][x] == 'S' :
            sx, sy = x, y
            flag = True
            break
    
    if flag :
        break


dist = [[-1] * W for _ in range(H)]
dist[sy][sx] = 0
q = deque([(sy, sx)])
while q :
    y, x = q.popleft()
    if dist[y][x] == K :
        continue
    
    for dx, dy in ((1, 0), (-1, 0), (0, 1), (0, -1)) :
        nx, ny = x + dx, y + dy
        if 0 <= nx < W and 0 <= ny < H and dist[ny][nx] == -1 and A[ny][nx] == '.' :
            dist[ny][nx] = dist[y][x] + 1
            q.append((ny, nx))

ret = float('inf')
for y in range(H) :
    for x in range(W) :
        if dist[y][x] < 0 :
            continue
        
        ret = min(ret, 1 + (min(H - y, y + 1, W - x, x + 1) - 1 + K - 1) // K)
        
print(ret)