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:
//In the Name of God
//Let Our Voices Echo
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define X real()
#define Y imag()
typedef long long ll;
typedef double ld; 
typedef complex<ld> point;
const ld eps=1e-9;
const int MAX=809,MOD=1e9+7;
int h,w,k,x,y,dis[MAX][MAX];
char ch[MAX][MAX];
queue<pair<int,int> > q;
int dx[]={0,+1,0,-1},dy[]={+1,0,-1,0};
int main()
{
	ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	memset(dis,-1,sizeof dis);
	cin>>h>>w>>k;
	for (int i=0;i<h;i++)
		for (int j=0;j<w;j++)
		{
			cin>>ch[i][j];
			if (ch[i][j]=='S') 
				x=i,y=j,ch[i][j]='.';
		}
	q.push({x,y});
	dis[x][y]=0;
	while (q.size())
	{
		int x=q.front().first,y=q.front().second;
		q.pop();
		if (dis[x][y]==k) continue;
		for (int i=0;i<4;i++)
		{
			int xx=x+dx[i],yy=y+dy[i];
			if (xx<0 || xx>=h || yy>=w || yy<0 || ch[xx][yy]=='#') continue;
			if (dis[xx][yy]==-1)
				dis[xx][yy]=dis[x][y]+1,q.push({xx,yy});
		}
	}
	int ans=1e9;
	for (int i=0;i<h;i++)
		for (int j=0;j<w;j++)
			if (dis[i][j]!=-1)
			{
				int d=min({i,h-i-1,j,w-j-1});
				if (d==0) ans=1;
				ans=min(ans,(d+k-1)/k+1);
			}
	cout<<ans;
	return 0;
}