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;
const int dx[4]={0,1,0,-1};
const int dy[4]={1,0,-1,0};
int n,m,k,mn,dist[805][805];
char s[805][805];
queue<pair<int,int> >q;
int dis(int x,int y){return (min(min(x,y),min(n-1-x,m-1-y))+k-1)/k;}
int main()
{
	scanf("%d%d%d",&n,&m,&k);
	memset(dist,-1,sizeof(dist));
	for(int i=0;i<n;i++)
	{
		scanf("%s",s[i]);
		for(int j=0;j<m;j++)if(s[i][j]=='S')
		{
			q.push(make_pair(i,j));
			dist[i][j]=0;
			mn=dis(i,j)+1;
		}
	}
	while(!q.empty())
	{
		int x=q.front().first,y=q.front().second;
		q.pop();
		if(dist[x][y]==k)break;
		for(int i=0;i<4;i++)
		{
			int xx=x+dx[i],yy=y+dy[i];
			if(xx>=0&&xx<n&&yy>=0&&yy<m&&s[xx][yy]=='.'&&dist[xx][yy]==-1)
			{
				dist[xx][yy]=dist[x][y]+1;
				q.push(make_pair(xx,yy));
				mn=min(mn,dis(xx,yy)+1);
			}
		}
	}
	printf("%d\n",mn);
	return 0;
}