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 N 805
using namespace std;
int n,m,k,a[N][N],sx,sy,dis[N][N];
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
char s[N];
pair<int,int>q[N*N];
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=n;i++)
	{
		scanf("%s",s+1);
		for (int j=1;j<=m;j++)
		{
			if (s[j]=='S') sx=i,sy=j;
			if (s[j]=='#') a[i][j]=1;
		}
	}
	int l=0,r=1,Ans=100000000;
	q[1]=make_pair(sx,sy);
	memset(dis,0x3f ,sizeof dis);
	dis[sx][sy]=0;
	while(l<r)
	{
		l++;
		int x=q[l].first,y=q[l].second;
	//	cout<<x<<' '<<y<<endl;
		Ans=min(Ans,min(min(x-1,n-x),min(y-1,m-y)));
		if (dis[x][y]<k)
		{
			for (int i=0;i<4;i++)
				if (x+dx[i]&&y+dy[i]&&x+dx[i]<=n&&y+dy[i]<=m&&dis[x+dx[i]][y+dy[i]]>10000000&&a[x+dx[i]][y+dy[i]]==0)
				{
					q[++r]=make_pair(x+dx[i],y+dy[i]);
					dis[x+dx[i]][y+dy[i]]=dis[x][y]+1;
				}
		}
	}
//	cout<<Ans<<endl;
	printf("%d\n",(Ans+k-1)/k+1);
}