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
h,w,k=map(int,input().split())
a=[list(input())for i in range(h)]
b=[[1]*w for i in range(h)]
c=0
for i in range(h):
    if c:
        break
    for j in range(w):
        if a[i][j]=="S":
            c=1
            p=[i,j]
            b[i][j]=0
def f(x,y):
    return min(x,w-y-1,y,h-x-1)
an=f(p[0],p[1])
q=[p]
c=0
while q and c!=k:
    c+=1
    d=[]
    x=[0,0,1,-1]
    y=[1,-1,0,0]
    for i,j in q:
        for s,t in zip(x,y):
            if 0<i+s<h-1 and 0<j+t<w-1:
                if b[i+s][j+t]:
                    if a[i+s][j+t]==".":
                        an=min(an,f(i+s,j+t))
                        b[i+s][j+t]=0
                        d.append([i+s,j+t])
            else:
                if a[i+s][j+t]==".":
                     print(1)
                     exit()
    q=d[::]
print(2+(an-1)//k)