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;
char Node[880][880];
int bs[880][880];
int zt[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
queue<int>QQ;
int my_min(int a,int b)
{
	return a<b ?a:b;
}
int merge(int r,int c)
{
	return r*800+c;
}
void ck(int sum,int &r,int &c)
{
	r=sum/800;
	c=sum%800;
}
int main()
{
	int h,w,k;
	scanf("%d %d %d",&h,&w,&k);
	memset(bs,-1,sizeof(bs));
	int sr,sc;
	for(int i=0;i<h;i++)
	{
		getchar();
		for(int k=0;k<w;k++)
		{
			scanf("%c",&Node[i][k]);
			if(Node[i][k]=='S')
				sr=i,sc=k;
		}
	}
	bs[sr][sc]=0;
	QQ.push(merge(sr,sc));
	int ans=1000000007;
	while(!QQ.empty())
	{
		int sum=QQ.front();
		QQ.pop();
		ck(sum,sr,sc);
	//	printf("%d %d\n",sr,sc);
		ans=my_min(ans,sr);
		ans=my_min(ans,sc);
		ans=my_min(ans,h-1-sr);
		ans=my_min(ans,w-1-sc);
		if(bs[sr][sc]==k)
			continue;
		for(int i=0;i<4;i++)
		{
			int nr,nc;
			nr=sr+zt[i][0];
			nc=sc+zt[i][1];
			if((nr>=0 && nr<h && nc>=0 && nc<w && bs[nr][nc]==-1 && Node[nr][nc]=='.'))
			{
				bs[nr][nc]=bs[sr][sc]+1;
				QQ.push(merge(nr,nc));
			}
		}
	}
	int aa=ans/k;
	if(ans%k!=0)
		aa++;
	printf("%d\n",aa+1);
	return 0;
}