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;
int n,m,k,sx,sy;
int head,tail;
int D[810][810],Q[1010000][2];
char p[810][810];
void Ins(int x,int y,int d)
{
	if(D[x][y]<=d||p[x][y]!='.')
	{
		return;
	}
	D[x][y]=d;
	++tail;
	Q[tail][0]=x;
	Q[tail][1]=y;
}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++)
	{
		scanf("%s",p[i]+1);
		for(int j=1;j<=m;j++)
		{
			if(p[i][j]=='S')
			{
				sx=i;
				sy=j;
				p[i][j]='.';
			}
			D[i][j]=1e9;
		}
	}
	Ins(sx,sy,0);
	while(head<tail)
	{
		++head;
		int x=Q[head][0];
		int y=Q[head][1];
		Ins(x-1,y,D[x][y]+1);
		Ins(x+1,y,D[x][y]+1);
		Ins(x,y-1,D[x][y]+1);
		Ins(x,y+1,D[x][y]+1);
	}
//	for(int i=1;i<=n;i++)
//	{
//		for(int j=1;j<=m;j++)
//		{
//			printf("%d ",D[i][j]);
//		}
//		printf("\n");
//	}
	int ans=1e9;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(D[i][j]<=k)
			{
				ans=min(ans,1+(min(min(i-1,n-i),min(j-1,m-j))+k-1)/k);
			}
		}
	}
	printf("%d",ans);
	return 0;
}