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:
H, W, K = map(int, input().split())
items = []

start = []
diff = float("inf")
p = []
prev = [[False] * W for _ in range(H)]
for i in range(H):
    items.append(list(input()))
    for j in range(W):
        s = items[i][j]
        if s == "S":
            start = [i, j]
            diff = min(diff, min(i, H - 1 - i, min(j, W - 1 - j)))
            p = [i, j]
            prev[i][j] = True

from collections import deque

que = deque([])

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

for i in range(4):
    x = dx[i]
    y = dy[i]
    if 0 <= start[0] + y <= H - 1 and 0 <= start[1] + x <= W - 1:
        tmp = items[start[0] + y][start[1] + x]
        if tmp == ".":
            que.append((start[0] + y, start[1] + x))
            prev[start[0] + y][start[1] + x] = True
k = K
while len(que) != 0:
    q = que.popleft()
    a = q[0]
    b = q[1]
    if abs(start[0] - a) + abs(start[1] - b) > K:
        continue
    c = min(min(a, H - 1 - a), min(b, W - 1 - b))
    if c < diff:
        diff = c
        p = (a, b)

    for i in range(4):
        x = dx[i]
        y = dy[i]
        if 0 <= a + y <= H - 1 and 0 <= b + x <= W - 1:
            if prev[a + y][b + x]:
                continue
            tmp = items[a + y][b + x]
            if tmp == ".":
                que.append((a + y, b + x))
                prev[a + y][b + x] = True

ans = 1
if diff == 0:
    print(ans)
else:
    print(ans + (K - 1 + diff) // K)