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<bits/stdc++.h>
using namespace std;
int n,m,k;
int mp[1005][1005],v[1005][1005];
char s[1005];
int sx,sy;
int k1[]={0,0,1,-1};
int k2[]={1,-1,0,0};
struct node
{
	int x,y;
	node(){;}
	node(int _x,int _y)
	{
		x=_x;y=_y;
	}
};
queue<node>q;
void bfs()
{
	q.push(node(sx,sy));
	int ans=0x3f3f3f3f;
	v[sx][sy]=0;
	while(!q.empty())
	{
		node tmp=q.front();q.pop();
		int step=min(min(tmp.x-1,tmp.y-1),min(n-tmp.x,m-tmp.y));
		if(!step)ans=1;
		else ans=min(ans,(step-1)/k+2);
		if(v[tmp.x][tmp.y]==k)continue;
		for(int i=0;i<=3;i++)
		{
			int xx=tmp.x+k1[i],yy=tmp.y+k2[i];
			if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&!mp[xx][yy]&&v[xx][yy]==-1)
			{
				v[xx][yy]=v[tmp.x][tmp.y]+1;
				q.push(node(xx,yy));
			}
		}
	}
	cout<<ans<<endl;
	return ;
}
int main()
{
	memset(v,-1,sizeof(v));
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",s+1);
		for(int j=1;j<=m;j++)
		{
			if(s[j]=='S')sx=i,sy=j;
			else if(s[j]=='#')mp[i][j]=1;
		}
	}
	bfs();
	return 0;
}