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
#include<bits/stdc++.h>
#define L long long
#define vi vector<int>
#define pb push_back
#define pi pair<int,int>
#define pii pair<pi,int>
#define aa first
#define bb second
#define xx aa.aa
#define yy aa.bb
#define zz bb
#define mp make_pair
#define mpp(a,b,c) mp(mp(a,b),c)
using namespace std;
int n,m,t,f[810][810],dx[4]={1,-1,0,0},dy[4]={0,0,1,-1},x[1000000],y[1000000],r,p=1e9;
char s[810][810];
inline void chk(int i,int j)
{
	int k=min(min(i-1,j-1),min(n-i,m-j));
	k=(k?((k-1)/t+1):0);
	p=min(p,k+1);
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j;
	scanf("%d%d%d",&n,&m,&t);
	for(i=1;i<=n;i++)
	  scanf("%s",&s[i][0]+1);
	for(i=1;i<=n;i++)
	  for(j=1;j<=m;j++)
	    if(s[i][j]=='S')
	      {
		   r=1;
		   x[1]=i;
		   y[1]=j;
		   f[i][j]=1;
		  }
	for(i=1;i<=r && f[x[i]][y[i]]<=t;i++)
	  for(j=0;j<4;j++)
	    if(x[i]+dx[j]>0 && x[i]+dx[j]<=n && y[i]+dy[j]>0 && y[i]+dy[j]<=m && !f[x[i]+dx[j]][y[i]+dy[j]] && s[x[i]+dx[j]][y[i]+dy[j]]=='.')
	      {
		   f[x[i]+dx[j]][y[i]+dy[j]]=f[x[i]][y[i]]+1;
		   x[++r]=x[i]+dx[j];
		   y[r]=y[i]+dy[j];
		  }
	for(i=1;i<=r;i++)
	  chk(x[i],y[i]);
	printf("%d\n",p);
	return 0;
}
