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 math
H,W,K=map(int,input().split())

A=[]
S=[0,0]
for i in range(H):
    a=list(input())
    for j in range(W):
        if a[j]=="S":
            S=[i,j]
    A.append(a)

#print(A)
#print(S)
import sys
input=sys.stdin.readline
B=[[1000 for j in range(W)]for i in range(H)]

B[S[0]][S[1]]=0
ans=min(S[0],S[1],W-1-S[1],H-1-S[0])
#print(ans)
Q=[[0,S[0],S[1]]]


for i in range(10**6):
    if i==len(Q):
        break
    else:
        for j in range(1):
            if Q[i][1]>0 and B[Q[i][1]-1][Q[i][2]]==1000 and B[Q[i][1]][Q[i][2]]<K and A[Q[i][1]-1][Q[i][2]]==".":
                Q.append([Q[i][0]+1,Q[i][1]-1,Q[i][2]])
                B[Q[i][1]-1][Q[i][2]]=B[Q[i][1]][Q[i][2]]+1
                if ans>min(Q[i][1]-1,Q[i][2],H-1-(Q[i][1]-1),W-1-(Q[i][2])):
                    ans=min(Q[i][1]-1,Q[i][2],H-1-(Q[i][1]-1),W-1-(Q[i][2]))
                    
            if Q[i][2]>0 and B[Q[i][1]][Q[i][2]-1]==1000 and B[Q[i][1]][Q[i][2]]<K and A[Q[i][1]][Q[i][2]-1]==".":
                Q.append([Q[i][0]+1,Q[i][1],Q[i][2]-1])
                B[Q[i][1]][Q[i][2]-1]=B[Q[i][1]][Q[i][2]]+1
                if ans>min(Q[i][1],Q[i][2]-1,H-1-(Q[i][1]),W-1-(Q[i][2]-1)):
                    ans=min(Q[i][1],Q[i][2]-1,H-1-(Q[i][1]),W-1-(Q[i][2]-1))
                    
            if Q[i][1]<H-1 and B[Q[i][1]+1][Q[i][2]]==1000 and B[Q[i][1]][Q[i][2]]<K and A[Q[i][1]+1][Q[i][2]]==".":
                Q.append([Q[i][0]+1,Q[i][1]+1,Q[i][2]])
                B[Q[i][1]+1][Q[i][2]]=B[Q[i][1]][Q[i][2]]+1
                if ans>min(Q[i][1]+1,Q[i][2],H-1-(Q[i][1]+1),W-1-(Q[i][2])):
                    ans=min(Q[i][1]+1,Q[i][2],H-1-(Q[i][1]+1),W-1-(Q[i][2]))
                    
            if Q[i][2]<W-1 and B[Q[i][1]][Q[i][2]+1]==1000 and B[Q[i][1]][Q[i][2]]<K and A[Q[i][1]][Q[i][2]+1]==".":
                Q.append([Q[i][0]+1,Q[i][1],Q[i][2]+1])
                B[Q[i][1]][Q[i][2]+1]=B[Q[i][1]][Q[i][2]]+1
                if ans>min(Q[i][1],Q[i][2]+1,H-1-(Q[i][1]),W-1-(Q[i][2]+1)):
                    ans=min(Q[i][1],Q[i][2]+1,H-1-(Q[i][1]),W-1-(Q[i][2]+1))
#print(B)
#print(ans)
print(math.ceil(ans/K)+1)