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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define N 1000
using namespace std;
char s[N][N];
int d[N][N],p[N*N][2],tail,ans,n,m,k,sx,sy,step[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
void bfs()
{
	memset(d,63,sizeof(d));
	p[1][0]=sx;p[1][1]=sy;tail=1;
	d[sx][sy]=0;
	for(int i=1;i<=tail;i++)
	{
		int x=p[i][0],y=p[i][1];
		for(int j=0;j<4;j++)
		{
			int xx=x+step[j][0],yy=y+step[j][1];
			if(xx>0 && xx<=n && yy>0 && yy<=m && d[xx][yy]>d[x][y]+1 && s[xx][yy]=='.')
			{
				d[xx][yy]=d[x][y]+1;
				p[++tail][0]=xx;p[tail][1]=yy;
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++) 
	{
		scanf("%s",s[i]+1);
		for(int j=1;j<=m;j++) if(s[i][j]=='S') sx=i,sy=j;
	}
	bfs();
	ans=999999999;
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			if(d[i][j]>k) continue;
			int t=min(min(i-1,n-i),min(j-1,m-j));
			if(t==0) {printf("1\n");return 0;}
			int tt=t/k;
			if(t%k) tt++;
			ans=min(ans,tt);
		}
	printf("%d\n",ans+1);
	return 0;
}