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>
#define pb push_back 
#define mp make_pair
using namespace std;

const int maxn=807;
typedef long long ll;

int n,m,ans,k,h,w,sx,sy;
int a[maxn][maxn];
char s[maxn][maxn];
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};

bool check(int x,int y)
{
	if(x>=1&&x<=h&&y>=1&&y<=w) return 1;
	return 0;
}

int dis(int x,int y)
{
	int ans=x-1;
	ans=min(ans,h-x);
	ans=min(ans,w-y);
	ans=min(ans,y-1);
	return ans;
}

void bfs(int x,int y)
{
	queue<pair<int,int> >q;
	q.push(mp(x,y));
	
	while(!q.empty())
	{
		int nx=q.front().first,ny=q.front().second;
		q.pop();
		if(a[nx][ny]==k) break;
		for(int i=0;i<4;++i)
		{
			int tx=nx+dx[i];
			int ty=ny+dy[i];
			if(check(tx,ty)&&a[tx][ty]==-1&&s[tx][ty]=='.')
			{
				a[tx][ty]=a[nx][ny]+1;
				q.push(mp(tx,ty));
			}
		}
	}
}

int main()
{
	int i,j;
	scanf("%d%d%d",&h,&w,&k);
	for(i=1;i<=h;++i)
	  scanf("%s",s[i]+1);
	
	for(i=1;i<=h;++i)
	  for(j=1;j<=w;++j)
	    if(s[i][j]=='S')
	      sx=i,sy=j;
	
	memset(a,-1,sizeof(a));
	
	a[sx][sy]=0;
	bfs(sx,sy);
	
	int dist=500;
	for(i=1;i<=h;++i)
	  for(j=1;j<=w;++j)
	    if(a[i][j]!=-1)
	    {
	    	dist=min(dist,dis(i,j));
	    }
	
	printf("%d",1+(dist+k-1)/k);
	return 0;
}