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:
h,w,k=map(int,input().split())
b=[]
v=[[0]*w for _ in range(h)]
sh=0
sw=0
for i in range(h):
    s=list(input())
    b.append(s)
    
    if 'S' in s:
        sh=i
        sw=s.index('S')
#ans=10**18
d=[[0,1],[1,0],[-1,0],[0,-1]]

def bfs(y,x):
    ans=10**18
    c=0
    q=[(y,x)]
    v[y][x]=1
    ans=min(y,x,h-1-y,w-1-x)
    while c<k:
        p=[]
        c+=1
        for i,j in q:
            for di,dj in d:
                if 0<=i+di<h and 0<=j+dj<w and v[i+di][j+dj]!=1 and b[i+di][j+dj]!='#':
                    p.append((i+di,j+dj))
                    v[i+di][j+dj]=1
                    ans=min(ans,i+di,j+dj,h-1-i-di,w-1-j-dj)
        q=p
    return ans
ans=bfs(sh,sw)
print(1+(ans+k-1)//k)