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())

maze = [[x for x in input()] for _ in range(h)]

sx = 0
sy = 0

for i in range(h):
    for j in range(w):
        if maze[i][j] == 'S':
            sx = j
            sy = i

d = [[float('inf')]*w for _ in range(h)]
d[sy][sx] = 0

q = deque([[sy,sx]])
c = 0

visited = [[False]*w for _ in range(h)]
visited[sy][sx] = True

while q:
    y,x = q.popleft()
    for j,k in [[1,0],[-1,0],[0,1],[0,-1]]:
        ny,nx = y+j,x+k
        if maze[ny][nx] == '.' and not visited[ny][nx]:
            d[ny][nx] = d[y][x] + 1
            if ny in {0,h-1} or nx in {0,w-1}:
                if d[ny][nx] <= K:
                    print(1)
                    exit()
                else:
                    continue
            else:
                visited[ny][nx] = True
                if d[ny][nx] == K:
                    continue
                else:
                    q.append([ny,nx])
    

def BFS(sy,sx):
    visited = [[-1]*w for _ in range(h)]
    q = deque([[sy,sx]])
    visited[sy][sx] = 0
    while q:
        y,x = q.popleft()
        if y == 0 or y == h-1 or x == 0 or x == w-1:
            return visited[y][x]
        for l,k in [[1,0],[-1,0],[0,1],[0,-1]]:
            new_y,new_x = y+l,x+k
            if visited[new_y][new_x] == -1:
                visited[new_y][new_x] = visited[y][x] + 1
                q.append([new_y,new_x])

ans = float('inf')

for i in range(h):
    for j in range(w):
        if d[i][j] <= K:
            ans = min(ans,min(i,h-1-i,j,w-1-j))

print((ans-1) // K +2)