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 <string.h>
#include <queue>
#include <algorithm>
using namespace std;
char mp[801][802];
int dist[801][801];
int main()
{
	int n,m,k;
	scanf("%d%d%d%*c",&n,&m,&k);
	int sx,sy;
	for (int i=1;i<=n;++i)
	{
		scanf("%s",mp[i]+1);
		for (int j=1;j<=m;++j)
		{
			if (mp[i][j]=='S')
			{
				sx=i;
				sy=j;
			}
		}
	}
	queue<pair<int,int> > q;
	q.push(make_pair(sx,sy));
	memset(dist,-1,sizeof(dist));
	dist[sx][sy]=0;
	while (!q.empty())
	{
		int fx=q.front().first,fy=q.front().second;
		q.pop();
		if (dist[fx][fy]==k)
			continue;
		for (int i=-1;i<=1;++i)
			for (int j=-1;j<=1;++j)
			{
				int ox=fx+i,oy=fy+j;
				if (ox>0&&ox<=n&&oy>0&&oy<=m&&(i==0||j==0)&&mp[ox][oy]!='#'&&dist[ox][oy]==-1)
				{
					dist[ox][oy]=dist[fx][fy]+1;
					q.push(make_pair(ox,oy));
				}
			}
	}
	int mindist=1000000000;
	for (int i=1;i<=n;++i)
		for (int j=1;j<=m;++j)
			if (dist[i][j]!=-1)
			{
				int curdist=min(i-1,min(n-i,min(j-1,m-j)));
				if (curdist<mindist)
					mindist=curdist;
			}
	printf("%d",1+(mindist%k?mindist/k+1:mindist/k));
	return 0;
}