Problem: 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

Solution (in PYTHON3):

Solution:
from collections import *
H,W,K=map(int,input().split())
A=[list(input()) for i in range(H)]
SX,SY=0,0
for i in range(H):
  for j in range(W):
    if A[i][j]=='S':
      SX,SY=j,i
D=deque()
Z=set()
P=1
def qao(y,x):
  if min(x,y)==0 or x+1==W or y+1==H:
    print(P)
    exit()
  if (y,x) in Z:
    return False
  D.append((y,x))
  Z.add((y,x))
  return True

w4x=(1,0,-1,0)
w4y=(0,1,0,-1)
def bfs(F):
  v=0
  for i in range(len(D)):
    v=D.popleft()
    for j in range(4):
      if A[v[0]+w4y[j]][v[1]+w4x[j]]=='.' or F:
        qao(v[0]+w4y[j],v[1]+w4x[j])

qao(SY,SX)
for i in range(K):
  bfs(0)
D=deque(Z)
while True:
  P+=1
  for i in range(K):
    bfs(1)