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<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define MAX 808
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int ans,n,m,K,sx,sy;
char g[MAX][MAX];
int Calc(int x,int y){return min(min(x-1,n-x),min(y-1,m-y));}
int dis[MAX][MAX];
int d[4][2]={1,0,0,1,-1,0,0,-1};
void BFS(int sx,int sy)
{
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)dis[i][j]=1e9;
	dis[sx][sy]=0;ans=1e9;
	queue<int> Qx,Qy;Qx.push(sx),Qy.push(sy);
	while(!Qx.empty())
	{
		int x=Qx.front(),y=Qy.front();Qx.pop();Qy.pop();
		ans=min(ans,(Calc(x,y)+K-1)/K);
		if(dis[x][y]>=K)continue;
		for(int i=0;i<4;++i)
		{
			int xx=x+d[i][0],yy=y+d[i][1];
			if(xx<1||xx>n||yy<1||yy>m)continue;
			if(g[xx][yy]=='#')continue;
			if(dis[xx][yy]<=dis[x][y]+1)continue;
			dis[xx][yy]=dis[x][y]+1;
			Qx.push(xx);Qy.push(yy);
		}
	}
}
int main()
{
	n=read();m=read();K=read();
	for(int i=1;i<=n;++i)scanf("%s",g[i]+1);
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			if(g[i][j]=='S')sx=i,sy=j;
	BFS(sx,sy);
	printf("%d\n",ans+1);
	return 0;
}