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 deque
import math

#入力
h,w,k=map(int,input().split())
a=[input() for i in range(h)]

#Sの位置の捜索
for i in range(h):
    for j in range(w):
        if a[i][j]=='S':
            sy,sx=i,j

#各種パラメータの初期設定
d=deque([[sy,sx]])
dist=[]
inf=-1
m=1000
for i in range(h):
    dist.append([inf]*w)
dist[sy][sx]=0

#一回目の魔法で移動できる範囲を幅優先探索
while len(d)!=0:
    b=d.popleft()
    p,q=b[0],b[1]
    for i in [[1,0],[0,-1],[-1,0],[0,1]]:
        by,bx=b[0]+i[0],b[1]+i[1]
        if bx>=0 and bx<w and by>=0 and by<h and a[by][bx]!='#' and dist[by][bx]==-1 and dist[p][q]<k:
            #訪問先が配列内かつ壁でなく未訪問、訪問元の距離がk以下のとき訪問
            d.append([by,bx])
            dist[by][bx]=dist[p][q]+1

#一回目で移動できる各地点からの魔法の使用回数を計算
for i in range(h):
    for j in range(w):
        if dist[i][j]!=-1:
            m=min(m,0--min(i,h-1-i,j,w-1-j)//k)

print(m+1)