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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};
char ch[1001];
int q[1000001][2],sx,sy,ans,b[1001][1001],i,j,k,n,m,a[1001][1001],w;
void bfs(){
	int x,y,tx,ty,i,j,l,r;
	for (i=1;i<=n;i++)
	 for (j=1;j<=m;j++) b[i][j]=-1;
	b[sx][sy]=0;l=0;r=1;q[1][0]=sx;q[1][1]=sy;
	while (l<r){
		x=q[++l][0];y=q[l][1];
		if (b[x][y]==w) return;
		for (i=0;i<4;i++){
			tx=x+dx[i];ty=y+dy[i];
			if (tx<1||tx>n||ty<1||ty>m||a[tx][ty]||b[tx][ty]!=-1) continue;
			b[tx][ty]=b[x][y]+1;q[++r][0]=tx;q[r][1]=ty;
		}
	}
}
int calc(int x){
	if (x==0) return 0;
	return (x-1)/w+1;
}
int main(){
	scanf("%d%d%d",&n,&m,&w);
	for (i=1;i<=n;i++){
		scanf("%s",ch);
		for (j=0;j<m;j++){
			if (ch[j]=='#') a[i][j+1]=1;else a[i][j+1]=0;
			if (ch[j]=='S') sx=i,sy=j+1;
		}
	}
	bfs();
	for (ans=n*m,i=1;i<=n;i++)
	 for (j=1;j<=m;j++)
	  if (b[i][j]!=-1){
	  	ans=min(ans,1+min(calc(min(i-1,n-i)),calc(min(j-1,m-j))));
	  }
	printf("%d\n",ans);
	return 0;
}