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
import sys
from collections import deque
input = sys.stdin.readline
inf = 10**8
DX = (-1, 0, 1, 0, -1, -1, 1, 1)
DY = (0, 1, 0, -1, -1, 1, -1, 1)
DX = DX[:4]
DY = DY[:4]


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


dist = [[inf]*W for _ in range(H)]
node = deque()
for i in range(H):
    for j in range(W):
        if grid[i][j] == "S":
            node.append((i, j))
            dist[i][j] = 0

# まず一回普通にBFS
memo = []
while node:
    x, y = node.popleft()
    memo.append((x, y))
    d = dist[x][y]
    if d >= K:
        continue
    for dx, dy in zip(DX, DY):
        nx = x + dx
        ny = y + dy
        if nx < 0 or ny < 0 or nx >= H or ny >= W:
            continue
        if grid[nx][ny] == "#":
            continue
        if dist[nx][ny] > d + 1:
            dist[nx][ny] = d + 1
            node.append((nx, ny))


ans = inf
for x, y in memo:
    d = min(x, y, H - x - 1, W - y - 1)
    ans = min(ans, 1 + (d + K - 1) // K)
print(ans)