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;
#define N 808
#define pii pair<int,int>
#define tri pair<pii,int>
#define mp make_pair
#define F first
#define S second 

int n,m,k,ans;
char s[N][N];
bool vis[N][N];
queue<tri> q;
bool inRange(int x,int y)
{
	return x>=0 && x<n && y>=0 && y<m && s[x][y]!='#';
}

int main()
{
	
	cin>>n>>m>>k;
	for(int i=0;i<n;i++)
	{
		cin>>s[i];
		for(int j=0;j<m;j++)
			if(s[i][j]=='S')
			{
				q.push(mp(mp(i,j),0));
				vis[i][j]=1;
			}
	}
	
	ans=1e9;
	while(!q.empty())
	{
		int x=q.front().F.F;
		int y=q.front().F.S;
		int step=q.front().S;
		q.pop();
		ans=min(ans,min(min(x,y),min(n-1-x,m-1-y)));
		for(int dx=-1;dx<=1;dx++)
		for(int dy=-1;dy<=1;dy++)
		{
			if(abs(dx)+abs(dy)!=1) continue;
			if(!inRange(x+dx,y+dy)) continue;
			if(vis[x+dx][y+dy]) continue;
			if(step==k) continue;
			vis[x+dx][y+dy]=1;
			q.push(mp(mp(x+dx,y+dy),step+1));
		}
	}
	cout<<(ans+k-1)/k+1<<'\n';
	
	return 0;
}