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
#include <bits/stdc++.h>
#define F first
#define S second
#define X real()
#define Y imag()
using namespace std;
typedef long long ll;
typedef long double ld;

const int MA=430;
string s[888];

int d[888][888];

int ru(int a, int b){
	int r=a/b;
	if (a%b) r++;
	return r;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int h,w,k;
	cin>>h>>w>>k;
	pair<int, int> sp;
	for (int i=0;i<h;i++){
		cin>>s[i];
		for (int ii=0;ii<w;ii++){
			d[i][ii]=(w*h+2)*2;
			if (s[i][ii]=='S'){
				d[i][ii]=0;
				sp={i, ii};
			}
		}
	}
	queue<pair<int, int> > bfs;
	bfs.push(sp);
	int ans=1e9;
	while (!bfs.empty()){
		auto x=bfs.front();
		bfs.pop();
		if (d[x.F][x.S]<=k){
			int dd=min(min(ru(x.F, k), ru(x.S, k)), min(ru(h-x.F-1, k), ru(w-x.S-1, k)));
			if (dd+1<ans){
				ans=dd+1;
			}
		}
		for (int j=-1;j<=1;j++){
			for (int jj=-1;jj<=1;jj++){
				if ((j==0)^(jj==0)){
					int ny=x.F+j;
					int nx=x.S+jj;
					if (ny<0||ny>=h) continue;
					if (nx<0||nx>=w) continue;
					if (s[ny][nx]=='#') continue;
					if (d[ny][nx]>d[x.F][x.S]+1){
						d[ny][nx]=d[x.F][x.S]+1;
						bfs.push({ny, nx});
					}
				}
			}
		}
	}
	cout<<ans<<endl;
}