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())
A = [[0 for _ in range(W)] for _ in range(H)]
for i in range(H):
    A_i_x = input()
    for j in range(W):
        A[i][j] = A_i_x[j]
        if A_i_x[j] == 'S':
            start_h = i
            start_w = j

dist_from_exit = [[0 for _ in range(W)] for _ in range(H)]
for i in range(H):
    for j in range(W):
        dist_from_exit[i][j] = min(i, j, H - 1 - i, W - 1 - j)

queue = deque()
queue.append([start_h, start_w])
visited = [[-1 for _ in range(W)] for _ in range(H)]
visited[start_h][start_w] = 0
dx = [1, 0, -1, 0]
dy = [0, 1, 0, -1]
dist_left_cand = [dist_from_exit[start_h][start_w]]
while queue:
    h, w = queue.pop()
    for i in range(4):
        if 0 <= h + dy[i] <= H-1 and 0 <= w + dx[i] <= W-1:
            if visited[h + dy[i]][w + dx[i]] == -1 and A[h + dy[i]][w + dx[i]] == '.' and visited[h][w] < K:
                queue.appendleft([h + dy[i], w + dx[i]])
                visited[h + dy[i]][w + dx[i]] = visited[h][w] + 1
                dist_left_cand.append(dist_from_exit[h + dy[i]][w + dx[i]])

dist_left_min = min(dist_left_cand)
print((dist_left_min - 1) // K + 1 + 1)
