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>
const int vx[4]={-1,0,0,1},vy[4]={0,-1,1,0};
int H,W,K;
char A[810][810];
struct Point{int x,y,d;}Q[810*810];
int min(int a,int b){return a<b?a:b;} 
int main(){
	scanf("%d%d%d",&H,&W,&K);
	for(int i=0;i<H;i++){
		scanf("%s",A[i]);
		for(int j=0;j<W;j++)if(A[i][j]=='S')*Q=(Point){i,j,0};
	}
	Point*h=Q,*t=Q+1;
	int ans=1<<30;
	for(;h<t;h++){
		ans=min(ans,1+(min(min(h->x,H-h->x-1),min(h->y,W-h->y-1))+K-1)/K);
		if(ans<2)break;
		if(h->d<K)for(int d=0,dx,dy;d<4;d++)if(A[dx=h->x+vx[d]][dy=h->y+vy[d]]=='.'){
			*t++=(Point){dx,dy,h->d+1};
			A[dx][dy]=1; 
		}
	}
	printf("%d\n",ans);
}