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

H,W,K = map(int,input().split())
src = [input() for i in range(H)]
sx = sy = None
for i,row in enumerate(src):
    if 'S' in row:
        sy = i
        sx = row.index('S')

dxy = [(0,1),(1,0),(0,-1),(-1,0)]
visited = [[0]*W for i in range(H)]
visited[sy][sx] = 1
q = deque([(0,sx,sy)])
mindist = H+W
while q:
    d,x,y = q.popleft()
    if d > K: break
    dist = min(x, W-x-1, y, H-y-1)
    mindist = min(dist, mindist)
    for dx,dy in dxy:
        nx,ny = x+dx,y+dy
        if not 0 <= nx < W: continue
        if not 0 <= ny < H: continue
        if src[ny][nx] == '#': continue
        if visited[ny][nx]: continue
        q.append((d+1,nx,ny))
        visited[ny][nx] = 1

ans = -(-mindist//K) + 1
print(ans)