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<map>
#include<queue>
using namespace std;
#define MAXN 810
#define MAXM 640010
#define INF 1000000000
#define PII pair<int,int>
char s[MAXN][MAXN];
int n,m,K,sx,sy,ans=INF;
int dis[MAXN][MAXN];
int d[]={1,0,-1,0,1};
queue<PII>Q;
void BFS()
{
	int l=0,r=0;
	dis[sx][sy]=K+1;
	Q.push(PII(sx,sy));
	while(!Q.empty())
	{
		PII p=Q.front();Q.pop();
		int nx=p.first,ny=p.second;
		ans=min(ans,min(nx-1,min(ny-1,min(n-nx,m-ny))));
		if(dis[nx][ny]==1) continue;
		for(int i=0;i<4;i++)
		{
			int px=nx+d[i],py=ny+d[i+1];
			if(px==0||py==0||px>n||py>m||s[px][py]=='#'||dis[px][py]!=0) continue;
			dis[px][py]=dis[nx][ny]-1;
			Q.push(PII(px,py));
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&K);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s[i]+1);
		for(int j=1;j<=m;j++)
			if(s[i][j]=='S')
				sx=i,sy=j;
	}
	if(sx==1||sx==n||sy==1||sy==m)
	{
		printf("0\n");
		return 0;
	}
	BFS();
	ans=(ans+K-1)/K+1;
	printf("%d\n",ans);
}