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<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<string>


using namespace std;


int h,w,d[810][810],k,sx,sy,ans=1e9;
queue<pair<int,int> > p;
string s[800];

int dx[] ={1,-1,0,0},dy[]={0,0,1,-1};


int main()
{
	cin >> h >> w >> k;
	for(int i = 0;i<h;i++)
	{
		cin >> s[i];
		for(int j=0;j<w;j++)
		{
			d[i][j] = -1;
			if(s[i][j] == 'S') sx = i,sy = j;
		}
	}
	p.push(make_pair(sx,sy));
	d[sx][sy] = k;
	while(!p.empty())
	{
		int x =p.front().first,y = p.front().second;
		p.pop();
		ans = min(ans,min(min(x,y),min(h-x-1,w-y-1)));
		if(x==0||y==0||x==h-1||y==w-1)ans=0;
		else for(int r = 0;r<4;r++)
		{
			int tx=x+dx[r],ty=y+dy[r];
			if((s[tx][ty]!='#')&&(d[tx][ty]<d[x][y]-1))
			{
				d[tx][ty]=d[x][y]-1;
				p.push(make_pair(tx,ty));
			}
		}
	}
	cout << (ans + k - 1)/k+1<<endl;
	return 0;
}