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<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int h,w,K,ans=1<<30,i,j,k,qx[805*805],qy[805*805],bg,ed,f[10][5]={{-1,0},{1,0},{0,-1},{0,1}},dis[805][805];
char s[805][805];
bool v[805][805];
int main()
{
	scanf("%d%d%d",&h,&w,&K);
	for(i=1;i<=h;++i)
	{
		scanf("%s",s[i]+1);
		for(j=1;j<=w;++j)
			dis[i][j]=1000000;
	}
	for(i=1;i<=h;++i)
	{
		for(j=1;j<=w;++j)
			if(s[i][j]=='S')
				break;
		if(j<=w)
			break;
	}
	bg=ed=1;
	qx[1]=i,qy[1]=j;
	v[i][j]=true;
	dis[i][j]=0;
	while(bg<=ed)
	{
		for(i=0;i<4;++i)
		{
			j=qx[bg]+f[i][0],k=qy[bg]+f[i][1];
			if(j>=1&&j<=h&&k>=1&&k<=w&&s[j][k]!='#'&&!v[j][k])
			{
				v[j][k]=true;
				++ed;
				qx[ed]=j,qy[ed]=k;
				dis[j][k]=dis[qx[bg]][qy[bg]]+1;
			}
		}
		++bg;
	}
	for(i=1;i<=h;++i)
		for(j=1;j<=w;++j)
			if(s[i][j]!='#'&&dis[i][j]<=K)
			{
				int g=min(min(i-1,j-1),min(h-i,w-j));
				if(g==0)
					ans=1;
				else
					ans=min(ans,(g-1)/K+2);
			}
	printf("%d",ans);
	return 0;
}