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 <string>
    #include <set>  
    #include <algorithm>
    #include <vector>
    #include <cmath>
    #include <queue>
    #include <deque>
    #include <cstring>
    #include <cstdio>
    #include <map>
    #include <numeric>
    #include <cassert>
    #include <iomanip>
    #include <sstream>
using namespace std;
const int dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
char a[808][808];
int h,w,k;
int dis[808][808];
queue<pair<int,int> > Q;
int main()
{
	cin>>h>>w>>k;
	memset(dis,-1,sizeof(dis));
	for(int i=0;i<h;i++)
		for(int j=0;j<w;j++)
		{
			cin>>a[i][j];
			if(a[i][j]=='S')
			{
				Q.push(make_pair(i,j));
				dis[i][j]=0;
			}
		}
	while(!Q.empty())
	{
		pair<int,int> x=Q.front();
		Q.pop();
		for(int i=0;i<4;i++)
		{
			int xx=x.first+dx[i],yy=x.second+dy[i];
			if(xx<0||yy<0||xx>=h||yy>=w)continue;
			if(a[xx][yy]=='.'&&dis[xx][yy]==-1)
			{
				Q.push(make_pair(xx,yy));
				dis[xx][yy]=dis[x.first][x.second]+1;
			}
		}
	}
	int ans=1e9+7;
	for(int i=0;i<h;i++)
		for(int j=0;j<w;j++)
			if(dis[i][j]!=-1&&dis[i][j]<=k)
			{
				int mn=min(min(i,j),min(h-i-1,w-j-1));
				ans=min(ans,(mn+k-1)/k+1);
			}
	cout<<ans<<endl;
	return 0;
}