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 <queue>
#include <iostream>
using namespace std;
const int INF=0x3f3f3f3f,maxn=804;
queue<int>q;
int a[maxn][maxn],n,m,k,vis[maxn][maxn],sx,sy,mina;
int lrud[5]={0,1,0,-1,0};
int Get(int x,int y)
{
	int X=(x+k-2)/k;
	X=min(X,(y+k-2)/k);
	X=min(X,(n-x+k-1)/k);
	X=min(X,(m-y+k-1)/k);
	return X;
}
void bfs()
{
	int i;
	vis[sx][sy]=0;
	q.push(sx);q.push(sy);
	while(!q.empty())
	{
		int x=q.front();q.pop();int y=q.front();q.pop();
	
		if(vis[x][y]==k)
			continue;
		for(i=1;i<=4;i++)
		{
			int gx=x+lrud[i-1];
			int gy=y+lrud[i];
			if(gx<1 || gy<1 || gx>n || gy>m || vis[x][y]+1>=vis[gx][gy] || a[gx][gy]==1)
				continue;
			vis[gx][gy]=vis[x][y]+1;
			q.push(gx);q.push(gy);

		}
	}
}
int main()
{
	int i,j,ans=0;
	cin>>n>>m>>k;
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=m;j++)
		{
			vis[i][j]=INF;
			char c;
			cin>>c;
			if(c=='.')
				a[i][j]=0;
			else
				if(c=='#')
					a[i][j]=1;
				else
				{
					sx=i,sy=j;
					a[i][j]=0;
				}
		}
	}
	mina=INF;
	bfs();
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=m;j++)
		{
			if(vis[i][j]!=INF)
			{	mina=min(mina,Get(i,j)+1);
			}
		}
	}
	cout<<mina;
	return 0;
}