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<bits/stdc++.h>
#define N 809
#define M 650009
using namespace std;

int m,n,lim,tail,d[N][N],h[M][2]; char s[N][N];
int calc(int x){ return (x)?(x-1)/lim+1:0; }
void ext(int x,int y,int z){
	if (x>0 && x<=m && y>0 && y<=n && s[x][y]=='.' && d[x][y]==-1){
		d[x][y]=z;
		h[++tail][0]=x; h[tail][1]=y;
	}
}
int main(){
	scanf("%d%d%d",&m,&n,&lim);
	int i,j,x,y,z,head=0; tail=1;
	memset(d,-1,sizeof(d));
	for (i=1; i<=m; i++){
		scanf("%s",s[i]+1);
		for (j=1; j<=n; j++) if (s[i][j]=='S'){
			s[i][j]='#';
			h[1][0]=i; h[1][1]=j; d[i][j]=0;
		}
	}
	int ans=1000000000;
	while (head<tail){
		x=h[++head][0]; y=h[head][1]; z=d[x][y]+1;
		if (d[x][y]>lim) continue;
		ans=min(ans,calc(min(min(x-1,m-x),min(y-1,n-y))));
		ext(x-1,y,z); ext(x+1,y,z);
		ext(x,y-1,z); ext(x,y+1,z);
	}
	printf("%d\n",ans+1);
	return 0;
}
