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
#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int dx[]={-1,0,1,0};
int dy[]={0,1,0,-1};
char ma[888][888];
int n,m,k,dist[888][888],pis[888][888],mn;
queue<pair<int,int> > q;
bool In(int x,int y)
{
	return (x>=0 && x<n && y>=0 && y<m);
}
int main()
{
	mn=1e9;
	scanf("%d%d%d",&n,&m,&k);
	for (int i=0;i<n;i++)
	scanf("%s",ma[i]);
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			pis[i][j]=min(min(i,j),min(n-i-1,m-j-1));
			pis[i][j]=(pis[i][j]+k-1)/k;
		}
	}
	memset(dist,-1,sizeof(dist));
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<m;j++)
		{
			if (ma[i][j]=='S') 
			{
				q.push(make_pair(i,j));
				dist[i][j]=0;
				mn=min(mn,pis[i][j]+1);
			}
		}
	}
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;
		q.pop();
		if (dist[x][y]==k) break;
		for (int i=0;i<4;i++)
		{
			int xx=x+dx[i],yy=y+dy[i];
			if (In(xx,yy) && ma[xx][yy]=='.' && dist[xx][yy]==-1)
			{
				mn=min(mn,pis[xx][yy]+1);
				dist[xx][yy]=dist[x][y]+1;
				q.push(make_pair(xx,yy));
			}
		}
	}
	printf("%d\n",mn);
	return 0;
}