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
sys.setrecursionlimit(2147483647)
INF=float("inf")
MOD=10**9+7
input=lambda:sys.stdin.readline().rstrip()
from itertools import product
from collections import deque
def resolve():
    h,w,k=map(int,input().split())
    G=[list(input()) for _ in range(h)]
    D=[(1,0),(-1,0),(0,1),(0,-1)]
    for i,j in product(range(h),range(w)):
        if(G[i][j]=='S'):
            i0,j0=i,j
            break

    dist=[[INF]*w for _ in range(h)]
    dist[i0][j0]=0
    # BFS
    Q=deque([(i0,j0)])
    while(Q):
        i,j=Q.popleft()
        for di,dj in D:
            ni=i+di; nj=j+dj
            if(not (0<=ni<h and 0<=nj<w)): continue
            if(dist[ni][nj]!=INF or G[ni][nj]=='#'): continue
            dist[ni][nj]=dist[i][j]+1
            Q.append((ni,nj))

    # 後は距離が k 以下の点から一直線に外に向かえばよい
    ans=INF
    for i,j in product(range(h),range(w)):
        if(dist[i][j]<=k):
            d=min(i,j,h-i-1,w-j-1)
            if(d==0):
                print(1)
                return
            else:
                ans=min(ans,2+(d-1)//k)
    print(ans)
resolve()