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<queue>
using namespace std;
const int INF=2e9+3;
char c[805][805];
int n,m,k;
int a[805][805];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int sx,sy;
bool ok(int x,int y)
{
	return x>=0&&y>=0&&x<n&&y<m;
}
int main()
{
	cin>>n>>m>>k;
	for(int i=0;i<n;i++)
	    for(int j=0;j<m;j++)
	    {
	    	cin>>c[i][j];
	    	if(c[i][j]=='S')
	    	    sx=i,sy=j;
		}
	for(int i=0;i<n;i++)
	    for(int j=0;j<m;j++)
	        a[i][j]=INF;
	a[sx][sy]=0;
	queue <pair<int,int> > q;
	q.push(make_pair(sx,sy));
	while(q.size())
	{
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		for(int i=0;i<4;i++)
		{
			int nx=x+dx[i];
			int ny=y+dy[i];
			if(ok(nx,ny)&&a[nx][ny]>a[x][y]+1&&c[nx][ny]!='#')
			{
				a[nx][ny]=a[x][y]+1;
				q.push(make_pair(nx,ny));
			}
		}
	}
	int ans=INF;
	for(int i=0;i<n;i++)
	    for(int j=0;j<m;j++)
	        if(a[i][j]<=k)
	        {
	        	int mn=min(min(i,j),min(n-i-1,m-j-1));
	        	int tmp=(k+mn-1)/k;
	        	ans=min(ans,tmp+1);
			}
	cout<<ans;
	return 0;
}