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
# coding: utf-8
# Your code here!
import heapq as hq
H,W,K = map(int,input().split())
A = []
X,Y=0,1
start = None
for h in range(H):
    tmp = list(input())
    A.append(tmp)
    if "S" in tmp:
        start = (tmp.index("S"), h)

def minCost(x,y):
    return min(start[X], W-1-start[X], start[Y], H-1-start[Y]) // K
search = []
searched = [[False for w in range(W)] for h in range(H)]
hq.heapify(search)
search.append((0, start[X], start[Y]))
vector = [(1,0),(-1,0),(0,1),(0,-1)]
max_x,min_x,max_y,min_y = 0,W-1,0,H-1
while len(search) > 0:
    cnt,x,y = hq.heappop(search)
    if x == 0 or x == W-1 or y == 0 or y == H-1:
        print(1)
        exit()
    if searched[y][x]:
        continue
    searched[y][x] = True
    max_x = max(max_x, x)
    min_x = min(min_x, x)
    max_y = max(max_y, y)
    min_y = min(min_y, y)
    for v in vector:
        newX,newY = x+v[X], y+v[Y]
        if not searched[newY][newX] and cnt < K and A[newY][newX] != "#":
            hq.heappush(search,(cnt+1,newX,newY))
dist = min(min_x, min_y, W-1-max_x, H-1-max_y)
print(1 + (dist+K-1)//K)