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())
S = []
for _ in range(H):
    S.append(list(input()))

D = [[1 for i in range(W)] for j in range(H)]

for x in range(H):
    for y in range(W):
        if S[x][y] == 'S':
            sx = x
            sy = y

D[sx][sy] = 0

dx = [1, 0, -1, 0]
dy = [0, 1, 0, -1]

ans = float('inf')

q = deque([[sx, sy, 0]])

while len(q) != 0:
    x, y, cnt = q.popleft()
    if cnt > K:
        break
    tmp = min(x, y , H - x - 1, W - y - 1)
    if tmp != 0:
        tmp = (tmp - 1) // K
        tmp += 2
    else:
        tmp = 1
    ans = min(ans, tmp)
    for i in range(4):
        nx = x + dx[i]
        ny = y + dy[i]
        if 0 <= nx < H and 0 <= ny < W and S[nx][ny] != '#' and D[nx][ny]:
            D[nx][ny] = 0
            q.append([nx, ny, cnt + 1])

print(ans)