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 CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
const int N = 810;
int h, w, k, ans, sx, sy;
int d[N][N], q[N * N * 2];
char Map[N][N];
int main()
{
	scanf("%d%d%d", &h, &w, &k);
	for(int i = 0; i <= h + 1; ++i)
		for(int j = 0; j <= w + 1; ++j) Map[i][j] = '#';
	for(int i = 1; i <= h; ++i)
	{
		char s[N];
		scanf("%s", s + 1);
		for(int j = 1; j <= strlen(s + 1); ++j) 
		{
			if(s[j] == 'S')
			{
				sx = i;
				sy = j;
			}
			Map[i][j] = s[j];
		}
	}
	ans = 1 << 29;
	int l = 1, r = 0;
	q[++r] = sx;
	q[++r] = sy;
	while(l <= r)
	{
		int x = q[l++], y = q[l++];
		ans = min(ans, min(min(x - 1, h - x), min(y - 1, w - y)));
		if(d[x][y] == k) continue;
		for(int i = 0; i < 4; ++i)
		{
			int xx = x + dx[i], yy = y + dy[i];
			if(!d[xx][yy] && Map[xx][yy] == '.') 
			{
				q[++r] = xx;
				q[++r] = yy;
				d[xx][yy] = d[x][y] + 1;
			}
		}
	}
	ans = (ans + k - 1) / k + 1;
	printf("%d\n", ans);		
	return 0;
}