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 <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
const int fx[4]={1,0,-1,0},fy[4]={0,1,0,-1};
int qx[640005],qy[640005],d[805][805],h,w,k;
char s[805][805];
int f(int x,int y){
	return (min(x-1,min(h-x,min(y-1,w-y)))+k-1)/k;
}
int main(){
	scanf("%d%d%d",&h,&w,&k);
	for (int i=1;i<=h;i++){
		scanf("%s",s[i]+1);
		for (int j=1;j<=w;j++)
			if (s[i][j]=='S') qx[0]=i,qy[0]=j;
	}
	memset(d,-1,sizeof d);
	int head=0,tail=1,ans=f(qx[0],qy[0]);
	for (d[qx[0]][qy[0]]=0;head!=tail && ans;head++){
		int x=qx[head],y=qy[head];
		if (d[x][y]==k) break;
		for (int i=0;i<4;i++){
			int x2=x+fx[i],y2=y+fy[i];
			if (d[x2][y2]==-1 && s[x2][y2]=='.'){
				d[qx[tail]=x2][qy[tail]=y2]=d[x][y]+1;
				tail++;
				ans=min(ans,f(x2,y2));
			}
		}
	}
	printf("%d\n",ans+1);
}