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>
using namespace std;
char c[880][880];
int n,m,k,x,y,dis[880][880],mm=1e9;
int dx[]={1,-1,0,0},dy[]={0,0,1,-1};
int main()
{
	cin>>n>>m>>k;
	for(int i=0;i<n;i++)
	{
		cin>>c[i];
		for(int j=0;j<m;j++)
			if(c[i][j]=='S')
			{
				x=i;
				y=j;
			}
	}
	queue<pair<int,int> >que;
	que.push(make_pair(x,y));
	memset(dis,-1,sizeof(dis));
	dis[x][y]=0;
	while(!que.empty())
	{
		int a=que.front().first;
		int b=que.front().second;
		que.pop();
		if(dis[a][b]>k)
			break;
		if(a==0||b==0||a==n-1||b==m-1)
		{
			cout<<1<<endl;
			return 0;
		}
		mm=min(mm,min(min(a,n-a-1),min(b,m-b-1)));
		for(int i=0;i<4;i++)
		{
			int u=a+dx[i],v=b+dy[i];
			if(c[u][v]!='#'&&dis[u][v]==-1)
			{
				dis[u][v]=dis[a][b]+1;
				que.push(make_pair(u,v));
			}
		}
	}
	cout<<(mm+k-1)/k+1<<endl;
	return 0;
}