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>
#include<cstring>
using namespace std;
int h,w,d[1000][1000],k,x,y,ans=1<<30,dx[]={1,-1,0,0},dy[]={0,0,1,-1};
queue<pair<int,int> > q;
char s[1000][1000];
int main()
{
	scanf("%d%d%d",&h,&w,&k);
	memset(d,-1,sizeof d);
	for(int i=0;i<h;i++)
	{
		scanf("%s",s[i]);
		for(int j=0;j<w;j++)
			if(s[i][j]=='S')x=i,y=j;
	}
	q.push(make_pair(x,y));
	d[x][y]=k;
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;q.pop();
		ans=min(ans,min(min(x,y),min(h-x-1,w-y-1)));
		if(!x||!y||x==h-1||y==w-1)ans=0;
		else for(int r=0;r<4;r++)
		{
			int tx=x+dx[r],ty=y+dy[r];
			if(s[tx][ty]!='#'&&d[tx][ty]<d[x][y]-1)
			{
				d[tx][ty]=d[x][y]-1;
				q.push(make_pair(tx,ty));
			}
		}
	}
	printf("%d\n",(ans+k-1)/k+1);
	return 0;
}