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>
#define INF 0x3f3f3f3f
using namespace std;
const int dx[4]={1,-1,0,0};
const int dy[4]={0,0,1,-1};
int n,m,k,d[801][801],xs,ys,ans=INF;
char s[801][801];
int asd(int i,int j)
{
	int r=i;
	r=min(r,j);
	r=min(r,n-1-i);
	r=min(r,m-1-j);
	return r;
}
queue<pair<int,int> > q;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>m>>k;
	for (int i=0;i<n;++i)
		for (int j=0;j<m;++j)
		{
			cin>>s[i][j];
			if (s[i][j]=='S') xs=i,ys=j;
		}
	memset(d,-1,sizeof(d));
	d[xs][ys]=0;
	q.push(make_pair(xs,ys));
	while (!q.empty())
	{
		int x=q.front().first;
		int y=q.front().second;
		q.pop();
		for (int i=0;i<4;++i)
		{
			int xx=x+dx[i];
			int yy=y+dy[i];
			if (xx>=0 && xx<n && yy>=0 && yy<m && s[xx][yy]=='.' && d[xx][yy]<0)
			{
				d[xx][yy]=d[x][y]+1;
				q.push(make_pair(xx,yy));
			}
		}
	}
	for (int i=0;i<n;++i)
		for (int j=0;j<m;++j)
			if (d[i][j]>=0 && d[i][j]<=k)
				ans=min(ans,1+(asd(i,j)+k-1)/k);
	cout<<ans;
	return 0;
}