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>
#define pb push_back
#define mp make_pair
#define fst first
#define snd second
#define fore(i,a,b) for(int i=a,ThxDem=b;i<ThxDem;++i)
using namespace std;
typedef long long ll;

char b[1024][1024];
int d[1024][1024];
int n,m,k;

bool valid(int i, int j){
	return i>=0&&i<n&&j>=0&&j<m&&b[i][j]!='#';
}

int asd(int i, int j){
	int r=i;
	r=min(r,j);
	r=min(r,n-1-i);
	r=min(r,m-1-j);
	return r;
}

queue<pair<int,int> > q;

int main(){
	scanf("%d%d%d",&n,&m,&k);
	int xs=-1,ys=-1;
	fore(i,0,n){
		scanf("%s",b[i]);
		fore(j,0,m)if(b[i][j]=='S')xs=i,ys=j;
	}
	memset(d,-1,sizeof(d));
	d[xs][ys]=0;
	q.push(mp(xs,ys));
	while(!q.empty()){
		int x=q.front().fst,y=q.front().snd;q.pop();
		fore(i,-1,2)fore(j,-1,2)if(abs(i)+abs(j)==1){
			int xx=x+i,yy=y+j;
			if(valid(xx,yy)&&d[xx][yy]<0){
				d[xx][yy]=d[x][y]+1;
				q.push(mp(xx,yy));
			}
		}
	}
	int r=1<<20;
	fore(i,0,n)fore(j,0,m)if(d[i][j]>=0&&d[i][j]<=k){
		r=min(r,1+(asd(i,j)+k-1)/k);
	}
	printf("%d\n",r);
	return 0;
}