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 <queue>

using namespace std;

const int maxn=810;

int vis[maxn][maxn],walkx[]={0,0,-1,1},walky[]={1,-1,0,0};
char str[maxn][maxn];

struct node
{
	int x,y,t;
};

queue <node> q;

int main()
{
	int n,m,k;
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",&str[i][1]);
		for(int j=1;j<=m;j++)
		{
			if(str[i][j]=='S')
			{
				q.push((node){i,j,0});
				vis[i][j]=1;
			}
		}
	}
	int ans=2e9;
	while(!q.empty())
	{
		node t=q.front();q.pop();
		int tt=min(t.x,t.y);
		ans=min(ans,(tt-1)/k+((tt-1)%k!=0));
		ans=min(ans,(n-t.x)/k+((n-t.x)%k!=0));
		ans=min(ans,(m-t.y)/k+((m-t.y)%k!=0));
		if(t.t==k)continue;
		for(int i=0;i<4;i++)
		{
			int newx=t.x+walkx[i],newy=t.y+walky[i];
			if(newx>0&&newx<=n&&newy>0&&newy<=m&&!vis[newx][newy]&&str[newx][newy]!='#')
			{
				q.push((node){newx,newy,t.t+1});
				vis[newx][newy]=1;
			}
		}
	}
	ans++;
	printf("%d\n",ans);
	return 0;
}