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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 800 + 7;
const int MOD = 1e9 + 7;

int dis[N][N];
pair<int, int> q[N*N];
string s[N];
int n, m, k;
int dr[] = {0, -1, 0, 1};
int dc[] = {1, 0, -1, 0};

bool valid(int r, int c)
{
	return r >= 0 && c >= 0 && r < n && c < m && s[r][c] != '#';
}

void bfs(int sr, int sc)
{
	memset(dis, 63, sizeof dis);
	int h = 0, t = 0;
	dis[sr][sc] = 0;
	q[t++] = {sr, sc};
	while (h < t)
	{
		int r = q[h].first;
		int c = q[h].second;
		h++;
		for (int i=0; i<4; i++)
		{
			int R = r + dr[i];
			int C = c + dc[i];
			if (valid(R, C) && dis[R][C] > dis[r][c] + 1)
			{
				dis[R][C] = dis[r][c] + 1;
				q[t++] = {R, C};
			}
		}
	}
}

int best(int r, int c)
{
	int A = min(r, n-1-r);
	int B = min(c, m-1-c);
	return (min(A, B) + k - 1) / k;
}

int main()
{
	ios::sync_with_stdio(false);
	
	memset(dis, 63, sizeof dis);


	cin >> n >> m >> k;
	for (int i=0; i<n; i++)
		cin >> s[i];
	for (int i=0; i<n; i++)
		for (int j=0; j<m; j++)
			if (s[i][j] == 'S')
				bfs(i, j);
	int ans = n*m;
	for (int i=0; i<n; i++)
		for (int j=0; j<m; j++)
			if (dis[i][j] <= k)
				ans = min(ans, best(i, j) + 1);
	cout << ans << "\n";
	return 0;
}