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 <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>
const int dx[4]={-1,1,0,0},dy[4]={0,0,1,-1};
typedef long long ll;
const int N=805;
int n,m,K,dis[N][N];
char mp[N][N];
using namespace std;

struct node
{
	int x,y;
};
queue<node> Q;
int main()
{
	#ifdef Kay
		freopen ("code.in","r",stdin);
		freopen ("code.out","w",stdout);
	#endif
	scanf ("%d %d %d",&n,&m,&K);
	int i,j;
	for (i=1;i<=n;i++) scanf ("\n%s",mp[i]+1);
	for (i=1;i<=n;i++)
	{
		for (j=1;j<=m;j++)
		if (mp[i][j]=='S') break;
		if (j<=m) break;
	}
	int s=i,t=j,nx,ny;
	memset(dis,-1,sizeof(dis));
	Q.push((node){i,j});dis[i][j]=0;
	while (!Q.empty())
	{
		node now=Q.front();Q.pop();
		for (i=0;i<4;i++)
		{
			nx=now.x+dx[i],ny=now.y+dy[i];
			if (nx<1 || ny<1 || nx>n || ny>m) continue;
			if (mp[nx][ny]=='#') continue;
			if (dis[nx][ny]==-1){dis[nx][ny]=dis[now.x][now.y]+1;Q.push((node){nx,ny});}
		}
	}
	int ans=n*m;
	for (i=1;i<=n;i++)
	for (j=1;j<=m;j++)
	if (mp[i][j]!='#' && dis[i][j]<=K && dis[i][j]!=-1)
	{
		if (i==1 || i==n || j==1 || j==m) ans=1;
		else ans=min(ans,(min(min(i-1,n-i),min(j-1,m-j))-1)/K+1+1);
	}
	cout<<ans;
	return 0;
}