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;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=800+5,inf=1e9,fx[4]={-1,0,1,0},fy[4]={0,-1,0,1};
char s[N][N];
int g[N][N];
pair<int,int>q[N*N];int tail;

int main()
{
	//freopen("1.in","r",stdin);
	int n,m,k;
	cin>>n>>m>>k;
	rep(i,1,n)scanf("%s",s[i]+1);
	rep(i,1,n)
	rep(j,1,m)
	if(s[i][j]!='#')g[i][j]=inf;
	rep(i,1,n)
	rep(j,1,m)
	if(s[i][j]=='S')
	{
		tail=1;
		q[1]=make_pair(i,j);
		g[i][j]=0;
	}
	int ans=inf;
	rep(head,1,tail)
	{
		int x=q[head].first,y=q[head].second;
		ans=min(ans,min(min(min(x-1,y-1),n-x),m-y));
		if(g[x][y]==k)continue;
		rep(i,0,3)
		{
			int tx=x+fx[i],ty=y+fy[i];
			if(g[tx][ty]==inf)
			{
				g[tx][ty]=g[x][y]+1;
				q[++tail]=make_pair(tx,ty);
			}
		}
	}
	cout<<(ans+k-1)/k+1;
}