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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int N=1010;
int n,m,k,x,y,dis[N][N];
char s[N][N];
typedef pair<int,int> pii;
queue<pii> Q;
const int fx[4]={1,0,-1,0},fy[4]={0,1,0,-1};
int main()
{
scanf("%d%d%d",&n,&m,&k);
for (int i=1;i<=n;i++) scanf("%s",s[i]+1);
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
if (s[i][j]=='S') x=i,y=j,s[i][j]='.';
for (int i=1;i<=n;i++)
for (int j=1;j<=m;j++)
dis[i][j]=1e9;
dis[x][y]=0;
int ans=1e9;
Q.push(pii(x,y));
while (!Q.empty()){
pii now=Q.front();Q.pop();
int x=now.first,y=now.second;
ans=min(ans,x-1);
ans=min(ans,y-1);
ans=min(ans,n-x);
ans=min(ans,m-y);
if (dis[x][y]==k) continue;
for (int i=0;i<4;i++){
int a=x+fx[i],b=y+fy[i];
if (s[a][b]!='.') continue;
if (dis[a][b]>dis[x][y]+1) dis[a][b]=dis[x][y]+1,Q.push(pii(a,b));
}
}
printf("%d\n",(ans+k-1)/k+1);
return 0;
}