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.Queue;
import java.util.Scanner;


public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt(), b = sc.nextInt(), k = sc.nextInt();
		char[][] cs = new char[a][b];
		int sx = -1, sy = -1;
		for(int i=0; i<a; i++) {
			String s = sc.next();
			if(s.indexOf('S') != -1) {
				sx = i;
				sy = s.indexOf('S');
			}
			cs[i] = s.toCharArray();
		}
		
		int[] dx = new int[]{0,0,1,-1};
		int[] dy = new int[]{1,-1,0,0};
		Queue<int[]> q = new LinkedList<int[]>();
		Queue<int[]> qq = new LinkedList<int[]>();
		q.add(new int[]{sx, sy});
		int min = get(cs, sx, sy);
		boolean[][] marked = new boolean[a][b];
		marked[sx][sy] = true;
		
		for(int j=0; j<k; j++) {
			for(int[] p : q) {
				for(int i=0; i<4; i++) {
					int newi=p[0]+dx[i];
					int newj=p[1]+dy[i];
					if(ok(cs, newi, newj) && !marked[newi][newj]) {
						marked[newi][newj] = true;
						qq.add(new int[]{newi, newj});
						int t = get(cs, newi, newj);
						if(t < min)	
							min = t;
					}
				}
			}
			q = qq;
			qq = new LinkedList<int[]>();
		}
		
		System.out.println(1 + (int)Math.ceil((0.0+min)/k));
	}

	private static int get(char[][] cs, int newi, int newj) {
		int row  = Math.min(newi, cs.length-1-newi);
		int col  = Math.min(newj, cs[0].length-1-newj);
		return Math.min(row, col);
	}

	private static boolean ok(char[][] cs, int newi, int newj) {
		if(newi<0 || newj<0 || newi>=cs.length || newj>=cs[0].length)	return false;
		if(cs[newi][newj] == '#')	return false;
		return true;
	}
}