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
def dist(r, c):
    return min(r, H - r - 1, c, W - c - 1)

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

for r in range(H):
    for c in range(W):
        if A[r][c] == 'S':
            sr = r
            sc = c

dxy = [(1, 0), (0, -1), (-1, 0), (0, 1)]
pos = [(sr, sc)]
dic = {(sr, sc): K}
cnt = 0
ans = H * W
i = 0
while i < len(pos):
    r0, c0 = pos[i]
    d0 = dist(r0, c0)
    ans = min(ans, (d0 + K - 1) // K + 1)
    m0 = dic[(r0, c0)]
    if 0 < m0:
        m = m0 - 1
        for (dx, dy) in dxy:
            r = r0 + dy
            c = c0 + dx
            d = dist(r, c)
            if 0 <= r < H and 0 <= c < W and A[r][c] != '#' and (r, c) not in dic:
                pos.append((r, c))
                dic[(r, c)] = m
    i += 1

print(ans)
