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.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int H = sc.nextInt();
		int W = sc.nextInt();
		int K = sc.nextInt();
		boolean[][] w = new boolean[H][W];
		int sy=0,sx=0;
		for(int i=0; i<H; i++) {
			char[] c = sc.next().toCharArray();
			for(int j=0; j<W; j++) {
				w[i][j] = c[j]=='#';
				if(c[j]=='S') {
					sy = i;
					sx = j;
				}
			}
		}

		int[] dy = {-1, 0, 1, 0};
		int[] dx = {0, -1, 0, 1};
		boolean[][] visited = new boolean[H][W];
		LinkedList<Long> q = new LinkedList<>();
		q.add(new Long(sy<<10 | sx));
		visited[sy][sx] = true;
		
		int nearest = 10000;
		
		while(!q.isEmpty()) {
			Long next = q.remove();
			int y = (int)((next>>10) & ((1<<10)-1));
			int x = (int)(next & ((1<<10)-1));
			int step = (int)(next>>20);
			
			int dis2Wall = Math.min(Math.min(y, H-1-y), Math.min(x, W-1-x));
			nearest = Math.min(nearest, dis2Wall);
			if(step<K) {
				for(int d=0; d<4; d++) {
					int ny = y + dy[d];
					int nx = x + dx[d];
					if(0<=ny && ny<H && 0<=nx && nx<W && !w[ny][nx] && !visited[ny][nx]) {
						q.add(new Long((step+1)<<20 | ny<<10 | nx));
						visited[ny][nx] = true;
					}
				}
			}
		}
		
		System.out.println((nearest+K-1)/K+1);
		sc.close();
	}
}