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<algorithm>
using namespace std;
const int N=805,M=N*N;
int dx[4]={0,0,1,-1},dy[4]={1,-1,0,0};
int n,m,k,i,j,a[N][N],h,t,qx[M],qy[M],ans;
char s[N];
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for (i=1;i<=n;i++){
		scanf("%s",s+1);
		for (j=1;j<=m;j++){
			if (s[j]=='S') qx[1]=i,qy[1]=j;
			if (s[j]=='#') a[i][j]=1;
		}
	}
	for (i=1;i<=n;i++) a[i][0]=a[i][m+1]=1;
	for (i=1;i<=m;i++) a[0][i]=a[n+1][i]=1;
	t=1;a[qx[1]][qy[1]]=1;ans=1e9;
	for (;h!=t;){
		h++;int x=qx[h],y=qy[h];
		ans=min(ans,min(min(x-1,n-x),min(y-1,m-y)));
		if (a[x][y]==k+1) continue;
		for (i=0;i<4;i++){
			int tx=x+dx[i],ty=y+dy[i];
			if (!a[tx][ty]){
				a[tx][ty]=a[x][y]+1;
				t++;qx[t]=tx;qy[t]=ty;
			}
		}
	}
	printf("%d",1+(ans+k-1)/k);
}