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<stdio.h>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
inline int read( )
{
  int sum=0;char c=getchar( );bool f=0;
  while(c<'0'||c>'9') {if(c=='-') f=1;c=getchar( );}
  while(c>='0'&&c<='9') {sum=sum*10+c-'0';c=getchar( );}
  if(f) return -sum;
  return sum;
}
const int N=805;
int n,m,K,qx[N*N],qy[N*N];
const int mx[ ]={1,-1,0,0};
const int my[ ]={0,0,1,-1};
int d[N][N];char s[N][N];
inline void BFS(int sx,int sy)
{
  int i,j,x,y,xx,yy,h=1,t=1,ans=1e9;
  qx[1]=sx;qy[1]=sy;d[sx][sy]=1;
  while(h<=t)
    {
      x=qx[h];y=qy[h++];
      if(d[x][y]>K) break;
      for(i=0;i<=3;i++)
	{
	  xx=x+mx[i];yy=y+my[i];
	  if(xx>n||xx<1||yy>m||yy<1||s[xx][yy]=='#'||d[xx][yy]) continue;
	  d[xx][yy]=d[x][y]+1;qx[++t]=xx;qy[t]=yy;
	}
    }
  for(i=1;i<=t;i++)
    {
      x=qx[i];y=qy[i];
      j=min(min(x-1,n-x),min(y-1,m-y));
      ans=min(ans,j);
    }
  printf("%d",(ans/K)+(ans%K!=0)+1);
}
int main( )
{
  int i,j;
  n=read( );m=read( );K=read( );
  for(i=1;i<=n;i++) scanf("%s",s[i]+1);
  for(i=1;i<=n;i++)
    for(j=1;j<=m;j++)
      if(s[i][j]=='S') {BFS(i,j);break;}
  return 0;
}