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 JAVA):

Solution:
import java.util.LinkedList;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner scan = new Scanner(System.in);
		int H = scan.nextInt();
		int W = scan.nextInt();
		int K = scan.nextInt();

		int[] dix ={-1,0,1,0};
		int[] diy ={0,1,0,-1};


		final int INF = 100000;

		LinkedList<Integer> nexti = new LinkedList<Integer>();
		LinkedList<Integer> nextj = new LinkedList<Integer>();

		int[][] A = new int[H][W];

		for(int i=0;i<H;i++){
			String S=scan.next();
			for(int j=0;j<W;j++){
				String s = S.substring(j,j+1);
				if(s.equals("S")){
					A[i][j]=0;
					nexti.add(i);
					nextj.add(j);
				}else if(s.equals(".")){
					A[i][j]=INF;
				}else{
					A[i][j]=-1;
				}
			}
		}




//MAP探索　1階目のループで1を記入
		for(int w=0;w<K;w++){
			int loop = nexti.size();
			for(int cnt=0;cnt<loop;cnt++){
				int i = nexti.poll();
				int j = nextj.poll();//次に探索できるi,j

				for(int x=0;x<4;x++){

						if(i+dix[x]>=0&&i+dix[x]<H&&j+diy[x]>=0&&j+diy[x]<W&&A[i+dix[x]][j+diy[x]]==INF){
							A[i+dix[x]][j+diy[x]] = w+1;
							nexti.add(i+dix[x]);
							nextj.add(j+diy[x]);
						}
				}
			}
		}

int min=INF;
//0以上K以下でゴールに最も近い点を探す
		for(int i=0;i<H;i++){
			for(int j=0;j<W;j++){
				if(A[i][j]>=0&&A[i][j]<INF){
					min = Math.min(min, i);
					min = Math.min(min, H-i-1);
					min = Math.min(min, j);
					min = Math.min(min, W-j-1);
				}
			}
			}

		int ans =1+(min+K-1)/K;



		System.out.println(ans);



	}

}