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<iostream>
#include<cstdlib>
#include<queue>
#include<cstdio>
using namespace std;
int v[4][2]={0,1,1,0,0,-1,-1,0};
bool ok[810][810];
int n,m,k;
int nx,ny;
bool vis[810][810];
int ans=1e9;
void bfs()
{
	queue<pair< pair<int,int > ,int > >q;
	q.push(make_pair(make_pair(nx,ny),0));
	vis[nx][ny]=true;
	while(!q.empty())
	{
		int x=q.front().first.first,y=q.front().first.second,w=q.front().second;
		q.pop();
		int res=x-1;
		res=min(res,n-x);
		res=min(res,y-1);
		res=min(res,m-y);
		if((res+k-1)/k+1<ans)ans=(res+k-1)/k+1;
		if(w==k)continue;
		for(int i=0;i<4;i++)
		{
			int xx=x+v[i][0],yy=y+v[i][1];
			if(xx>0 && xx<=n && yy>0 && yy<=m && ok[xx][yy] && !vis[xx][yy])
			{
				vis[xx][yy]=true;
				q.push(make_pair(make_pair(xx,yy),w+1));
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			char x;
			cin>>x;
			if(x=='.' || x=='S')ok[i][j]=true;
			if(x=='S')
			{
				nx=i;ny=j;
			} 
		}
	}
	bfs();
	printf("%d",ans);
	return 0;
}