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<cstdio>
#include<algorithm>
using namespace std;
#define MN 800
char s[MN+5][MN+5];
int qx[MN*MN+5],qy[MN*MN+5],qn,d[MN+5][MN+5];
const int o[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int main()
{
	int n,m,k,i,j,x,y,ans=MN;
	scanf("%d%d%d",&n,&m,&k);
	for(i=1;i<=n;++i)
	{
		scanf("%s",s[i]+1);
		for(j=1;j<=m;++j)if(s[i][j]=='S')d[qx[0]=i][qy[0]=j]=1;
	}
	for(i=0;i<=qn;++i)
	{
		ans=min(ans,(min(min(qx[i]-1,n-qx[i]),min(qy[i]-1,m-qy[i]))+k-1)/k);
		if(d[qx[i]][qy[i]]<=k)for(j=0;j<4;++j)
		{
			x=qx[i]+o[j][0];y=qy[i]+o[j][1];
			if(s[x][y]=='.'&&!d[x][y])d[x][y]=d[qx[i]][qy[i]]+1,qx[++qn]=x,qy[qn]=y;
		}
	}
	printf("%d",ans+1);
}