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<cstring>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN=805,fx[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int H,W,K;
int map[MAXN][MAXN],st[2];
bool vis[MAXN][MAXN];
struct node
{
	int x,y,t;
	node(){}
	node(int xx,int yy,int tt)
	{
		x=xx,y=yy,t=tt;
	}
};
queue<node>q;
int main()
{
	//freopen("room.in","r",stdin);
	//freopen("room.out","w",stdout);
	scanf("%d %d %d",&H,&W,&K);
	for(int i=1;i<=H;i++)
	{
		scanf("\n");
		for(int j=1;j<=W;j++)
		{
			char c;
			scanf("%c",&c);
			if(c=='.')
				map[i][j]=0;
			if(c=='#')
				map[i][j]=1;
			if(c=='S')
			{
				st[0]=i;
				st[1]=j;
			}
		}
	}
	if(st[0]==1||st[1]==1||st[0]==H||st[1]==W)
	{
		printf("0\n");
		return 0;
	}
	q.push(node(st[0],st[1],K));
	vis[st[0]][st[1]]=1;
	int ans=H*W;
	while(!q.empty())
	{
		node k=q.front();
		q.pop();
		int Mi=min(min(k.x-1,H-k.x),min(k.y-1,W-k.y));
		ans=min(1+(Mi+K-1)/K,ans);
		if(k.t==0)
			continue;
		for(int o=0;o<4;o++)
		{
			int xx=k.x+fx[o][0];
			int yy=k.y+fx[o][1];
			if(1<=xx&&xx<=H&&1<=yy&&yy<=W&&map[xx][yy]==0&&vis[xx][yy]==0)
			{
				vis[xx][yy]=1;
				q.push(node(xx,yy,k.t-1));
			}
		}
	}
	printf("%d\n",ans);
	fclose(stdin);
	fclose(stdout);
}