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
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__

h,w,k = map(int, input().split())
board = [list(input()) for i in range(h)]

for i in range(h):
    for j in range(w):
        if board[i][j] == "S":
            sx, sy = i, j

from collections import deque
dxdy = [(0,1),(1,0),(0,-1),(-1,0)]
def bfs(sx, sy):
    q = deque()
    q.append((sx, sy))
    used = [[0]*w for i in range(h)]
    ans = float("inf")
    while q:
        x,y = q.pop()
        if used[x][y]:
            continue
        used[x][y] = 1
        tmp = -(-min(x, h-1-x, y, w-1-y)//k)
        ans = min(ans, tmp+1)
        for dx, dy in dxdy:
            nx = x + dx
            ny = y + dy
            if abs(nx-sx)+abs(ny-sy)>k:
                continue
            if 0<=nx<h and 0<=ny<w and board[nx][ny] == ".":
                if nx == 0 or nx == h-1 or ny == 0 or ny == w-1:
                    return 1
                if not used[nx][ny]:
                    q.append((nx,ny))
    return ans

ans = bfs(sx, sy)
print(ans)