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 mp  make_pair
using namespace std;

typedef pair <int,int>  PII;
const int dx[4]={-1,1,0,0},dy[4]={0,0,-1,1};
int n,m,k,a[805][805],dis[805][805],ans;
char ch;
queue <PII> Q;

void work()
{
	scanf("%d %d %d",&n,&m,&k),ans=2*n;
	for (int i=1; i<=n; i++)
		for (int j=1; j<=m; j++)
			{
				do  ch=getchar();  while ((ch!='#')&&(ch!='.')&&(ch!='S'));
				if (ch!='#')  a[i][j]=1;
				if (ch=='S')  dis[i][j]=1,Q.push(mp(i,j));
			}
	while (!Q.empty())
		{
			int x=Q.front().first,y=Q.front().second,d;  Q.pop();
			d=min(min(x-1,n-x),min(y-1,m-y));
			ans=min(ans,(d+k-1)/k+1);
			if (dis[x][y]<=k)
				for (int i=0; i<4; i++)
					if ((a[x+dx[i]][y+dy[i]])&&(!dis[x+dx[i]][y+dy[i]]))
						dis[x+dx[i]][y+dy[i]]=dis[x][y]+1,Q.push(mp(x+dx[i],y+dy[i]));
		}
	printf("%d",ans);
}

int main()
{
	work();
	return 0;
}