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
input = sys.stdin.readline

import itertools

# 初日以外は何も気にせず動けるとしてよい。初日に到達可能な場所のうち、最もゴールに近いものはどこか

H,W,K = map(int,input().split())
A = '#' * (W+2)
for _ in range(H):
    A += '#' + input().rstrip() + '#'
A += '#' * (W+2)

start = A.find('S')
dxs = (1,-1,(W+2),-(W+2))

visited = set()
q = [start]
visited.add(start)
for _ in range(K):
    qq = []
    for x,dx in itertools.product(q,dxs):
        y = x+dx
        if A[y] == '#':
            continue
        if y in visited:
            continue
        visited.add(y)
        qq.append(y)
    q = qq

nearest = 10**9
for x in visited:
    row,col = divmod(x,W+2)
    d = min(row-1,H-row,col-1,W-col)
    if nearest > d:
        nearest = d

answer = 1 + (nearest+K-1)//K
print(answer)