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 <stdio.h>
#include <queue>
using namespace std;

int H,W,K; char map[808][808]; int dist[808][808];

int main()
{
	scanf ("%d %d %d",&H,&W,&K);
	queue<pair<int, int> > Q;
	for (int i=0;i<H;i++){
		scanf ("%s",map[i]);
		for (int j=0;j<W;j++){
			dist[i][j] = -1;
			if (map[i][j] == 'S'){
				Q.push({i,j}); dist[i][j] = 0;
			}
		}
	}

	while (!Q.empty()){
		int x = Q.front().first, y = Q.front().second; Q.pop();
		int dx[4] = {0,1,0,-1};
		int dy[4] = {1,0,-1,0};
		for (int k=0;k<4;k++){
			int px = x + dx[k], py = y + dy[k];
			if (px < 0 || px >= H || py < 0 || py >= W) continue;
			if (dist[px][py] == -1 && map[px][py] != '#'){
				Q.push({px,py}); dist[px][py] = dist[x][y] + 1;
			}
		}
	}

	int ans = 0x7fffffff;
	for (int i=0;i<H;i++) for (int j=0;j<W;j++) if (dist[i][j] != -1 && dist[i][j] <= K){
		int r = (min(min(i,H-i-1),min(j,W-j-1)) + K - 1) / K + 1;
		ans = min(r,ans);
	}
	printf ("%d\n",ans);

	return 0;
}