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:
//Love and Freedom.
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#define inf 20021225
#define ll long long
#define N 810
#define pa pair<int,int>
using namespace std;
int read()
{
	int f=1,s=0; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
	while(ch>='0'&&ch<='9')	s=s*10+ch-'0',ch=getchar();
	return f*s;
}
int n,m,k,dx[4]={0,0,-1,1},dy[4]={1,-1,0,0};
int F(int x,int y){return min(min((x-1+k-1)/k,(y-1+k-1)/k),min((n-x+k-1)/k,(m-y+k-1)/k));}
int ans=inf; bool vis[N][N]; int dis[N][N];
queue<pa> q;
char ch[N];
int main()
{
	n=read(),m=read(),k=read(); int x,y;
	for(int i=1;i<=n;i++)
	{
		scanf("%s",ch+1);
		for(int j=1;j<=m;j++)
			if(ch[j]=='#')	vis[i][j]=1;
			else if(ch[j]=='S') q.push(make_pair(i,j));
	}
	while(!q.empty())
	{
		pa tmp=q.front(); x=tmp.first,y=tmp.second; q.pop();
		ans=min(ans,F(x,y)+1);
		if(dis[x][y]==k)	continue;
		for(int i=0;i<4;i++)
		{
			int nx=x+dx[i],ny=y+dy[i];
			if(!nx||!ny||nx>n||ny>m||vis[nx][ny])	continue;
			vis[nx][ny]=1; dis[nx][ny]=dis[x][y]+1; q.push(make_pair(nx,ny));
		}
	}
	printf("%d\n",ans);
	return 0;
}