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 <cstdio>
#include <algorithm>
#include <queue>
#define maxn 805
#define INF 23333333
#define hh(x) (((x)>0)?((x)-1)/K+1:0)
using namespace std;

int ans=INF,sum,H,W,K,x1,y1,f[maxn][maxn][4];
int vis[maxn][maxn],fx[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
int tmp,xxx,yyy,x,y,dep;
char map[maxn][maxn];

struct xx
{
	int x,y,dep;
};
queue <xx > q;
inline int check(int x,int y)
{
	return 1<=x&&x<=H&&
	       1<=y&&y<=W&&
		   map[x][y]!='#';
}

inline void bfs()
{
	q.push((xx){x1,y1,0});
	while (!q.empty())
	{
		xx now=q.front();q.pop();
		x=now.x,y=now.y,dep=now.dep;
		if (dep>K||vis[x][y]) continue;
		vis[x][y]=1;
		tmp=min(min(H-x,W-y),min(x-1,y-1));
		tmp=(tmp+K-1)/K;
		if (tmp+1>ans) continue;
		ans=min(ans,tmp+1);
	//printf("-- x:%d y:%d dep:%d tmp+1:%d\n--",x,y,dep,tmp+1);
		for (int i=0;i<4;++i)
		{
			xxx=x+fx[i][0];
			yyy=y+fx[i][1];
			if (check(xxx,yyy))
				  q.push((xx){xxx,yyy,dep+1});
		}
	}
	return ;
}

int main()
{
	scanf("%d%d%d",&H,&W,&K);
	for (int i=1;i<=H;++i) scanf("%s",map[i]+1);
	for (int i=1;i<=H;++i)
	  for (int j=1;j<=W;++j)
	  	  if ('S'==map[i][j]) x1=i,y1=j;
	if (x1==1||x1==H||y1==1||y1==W) ans=0;
	else bfs();
	printf("%d",ans);
	return 0;
}