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 <cstdio>
#include <algorithm>
const int dx[4] = { 0, 0, 1, -1 }, dy[4] = { -1, 1, 0, 0 };
int N, M, K, a[802][802], X, Y, qx[640001], qy[640001], dis[802][802], O = 1000000000;
int main()
{
	scanf("%d%d%d", &N, &M, &K);
	for (int i = 1; i <= N; i++)
	{
		static char I[802];
		scanf("%s", I + 1);
		for (int j = 1; j <= M; j++)
			if (I[j] != '#')
			{
				a[i][j] = 1;
				if (I[j] == 'S')
				{
					X = i;
					Y = j;
				}
			}
	}
	int H = 0, T = 1;
	qx[1] = X;
	qy[1] = Y;
	dis[X][Y] = 1;
	while (H < T)
	{
		H++;
		int ux = qx[H], uy = qy[H];
		for (int i = 0; i < 4; i++)
		{
			int vx = ux + dx[i], vy = uy + dy[i];
			if (a[vx][vy] && !dis[vx][vy])
			{
				T++;
				qx[T] = vx;
				qy[T] = vy;
				dis[vx][vy] = dis[ux][uy] + 1;
			}
		}
	}
	for (int i = 1; i <= N; i++)
		for (int j = 1; j <= M; j++)
			if (1 <= dis[i][j] && dis[i][j] <= K + 1)
				O = std::min(O, (std::min(std::min(i - 1, N - i), std::min(j - 1, M - j)) + K - 1) / K + 1);
	printf("%d\n", O);
}
