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<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#define mp(x,y) make_pair(x,y)
using namespace std;

typedef pair<int,int> pi;

const int N=805;

int n,m,k,dis[N][N],sx,sy,dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
bool ma[N][N];
char str[N];
queue<pi> que;

void bfs()
{
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			dis[i][j]=k+1;
	dis[sx][sy]=0;que.push(mp(sx,sy));
	while (!que.empty())
	{
		int x=que.front().first,y=que.front().second;que.pop();
		if (dis[x][y]==k) break;
		for (int i=0;i<4;i++)
		{
			int p=x+dx[i],q=y+dy[i];
			if (p<1||p>n||q<1||q>m||ma[p][q]||dis[p][q]<=k) continue;
			dis[p][q]=dis[x][y]+1;que.push(mp(p,q));
		}
	}
}

int get(int x)
{
	return !x?0:(x-1)/k+1;
}

int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",str+1);
		for (int j=1;j<=m;j++)
			if (str[j]=='S') sx=i,sy=j;
			else if (str[j]=='#') ma[i][j]=1;
	}
	bfs();
	int ans=n*m;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=m;j++)
			if (dis[i][j]<=k) ans=min(ans,min(get(i-1),min(get(j-1),min(get(n-i),get(m-j))))+1);
	printf("%d",ans);
	return 0;
}