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<algorithm>
#include<queue>
#define MAXN 806
using namespace std;
int n,m,k,step[MAXN*MAXN],way[4][2]={{-1,0},{1,0},{0,1},{0,-1}},s,ans;
char a[MAXN][MAXN];
queue<int> q;
int main()
{
	//freopen("room.in","r",stdin);
	//freopen("room.out","w",stdout);
	scanf("%d%d%d",&n,&m,&k);
	ans=n*m*666;
	for(int i=1;i<=n;i++)
		scanf("%s",a[i]+1);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			if(a[i][j]=='S')
			{
				if(i==1||i==n||j==1||j==m)
				{
					printf("0");
					return 0;
				}
				s=(i-1)*m+j;
				break;
			}
	step[s]=1;
	q.push(s);
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		if(step[now]==k+1) continue;
		int nowx=(now-1)/m+1;
		int nowy=now-(nowx-1)*m;
		for(int i=0;i<4;i++)
		{
			int nex=nowx+way[i][0];
			int ney=nowy+way[i][1];
			int ne=(nex-1)*m+ney;
			if(step[ne]||a[nex][ney]!='.') continue;
			step[ne]=step[now]+1;
			q.push(ne);
		}
	}
	for(int i=1;i<=n*m;i++)
	{
		if(step[i])
		{
			int x=(i-1)/m+1;
			int y=i-(x-1)*m;
			int p=min(min(x-1,n-x),min(y-1,m-y));
			int q=p/k;
			if(p%k) q++;
			ans=min(ans,q+1);
		}
	}
	printf("%d",ans);
}