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;
const int INF=1e9,N=1000,dx[]={0,0,1,-1},dy[]={1,-1,0,0};
int n,m,k,x,y;
char c[N][N];
bool vis[N][N];
int main()
{
	//freopen("qc.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<n;i++)
	{
		scanf("%s",&c[i]);
		for(int j=0;j<m;j++)if(c[i][j]=='S')x=i,y=j;
	}
	int ans=min(min(x,y),min(n-x-1,m-y-1));
	vis[x][y]=1;
	if(ans%k==0)ans=ans/k;
	else ans=ans/k+1;
	int minv=ans+1;
	queue<pair<int,pair<int,int> > >q;
	q.push(make_pair(0,make_pair(x,y)));
	while(!q.empty())
	{
		pair<int,pair<int,int> >t=q.front();
		q.pop();
		if(t.first>=k)continue;
		for(int w=0;w<4;w++)
		{
			int nx=t.second.first+dx[w],ny=t.second.second+dy[w];
			if(nx<0||nx>=n||ny<0||ny>=m)continue;
			if(vis[nx][ny])continue;
			if(c[nx][ny]=='#')continue;
			vis[nx][ny]=true;
			int res=min(min(nx,ny),min(n-nx-1,m-ny-1));
			if(res%k==0)res=res/k;
			else res=res/k+1;
			minv=min(minv,res+1);
			q.push(make_pair(t.first+1,make_pair(nx,ny)));
		}
	}
	printf("%d\n",minv);
	return 0;
}