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<queue>
#include<cctype>
#include<cstdio>
using namespace std;
const int maxn=800,dir[4][2]={0,1,1,0,0,-1,-1,0};
bool a[maxn+10][maxn+10]; int stx,sty,n,m,k,ans=1e9;
struct node{int x,y,d;}; queue<node> q;
void work(int x){ans=min(ans,(x-1)/k+2);}
int main(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j){
			char c=getchar(); for(;!isgraph(c);c=getchar()); 
			if(c=='S') q.push((node){i,j,k}),a[i][j]=1;
			else if(c=='#') a[i][j]=1;
		}
	for(;!q.empty();q.pop()){
		node x=q.front(); 
		if(x.x==1||x.x==n||x.y==1||x.y==m) ans=1;
		else{
			work(x.x-1); work(n-x.x); work(x.y-1); work(m-x.y);
		}
		if(x.d) for(int i=0;i<4;++i){
			node e=(node){x.x+dir[i][0],x.y+dir[i][1],x.d-1};
			if(e.x<=0||e.x>n||e.y<=0||e.y>m||a[e.x][e.y]) continue;
			a[e.x][e.y]=1; q.push(e);
		}
	}
	printf("%d",ans); 
}